a. For each digit there are 7 choices and thus the number of possible codes is then:

$$
7cdot 7 cdot 7 cdot 7=7^4=2,401
$$

b. We need to make four decisions about the digits (one for each digit) and each digit has 7 choices.

c. There will be 4 branch points (one each for the digit) and 7 branches at each point. Note that these answers contain the same integers as the answers of Nick’s.

d. There are 7 choices for each decision and thus the number of possibilities are:

$$
7cdot 7 cdot 7 cdot 7=7^4=2,401
$$

e. The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
dfrac{1}{2,401}approx 0.0004=0.04%
$$

Multiplying the number of choices for each digit:

$$
7times 6times 5times 4=840
$$

Thus in total, there are 840 possible passwords if you can’t repeat numbers.

a. For the first digit there are 9 choices, the second 8 and the third 7:

$$
9cdot 8 cdot 7=504
$$

b. There are 9 choices for each digit:

$$
9cdot 9 cdot 9 =9^3=729
$$

c. The similarities is that the same digits are possible and the same number of digits are needed. The difference is that in (a) the digits cannot be repeated, while in (b) the same digits can be repeated.

a. For the first period there are 6 choices, the second 5, the third 4, etc.:

$$
6cdot 5 cdot 4cdot 3 cdot 2 cdot 1=720
$$

b. For the first period there is 1 choice, the second 5, the third 4, etc.:

$$
1cdot 5 cdot 4cdot 3 cdot 2 cdot 1=120
$$

c. The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
dfrac{120}{720}=dfrac{1}{6}approx 0.167=16.7%
$$

d. The number of possibilities of having first-period pre-calculus and second-period physics:

$$
1cdot 1 cdot 4 cdot 3 cdot 2cdot 1=24
$$

The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
dfrac{24}{720}=dfrac{1}{30}approx 0.033=3.3%
$$

a. 720

b. 720

c. $frac{1}{6}approx 0.167=16.7%$

d. $frac{1}{30}approx 0.033=3.3%$

a. $8!=40,320$, $7!=5,040$, $6!=720$, $5!=120$, $4!=24$, $3!=6$, $2!=2$ and $1!=1$

b. 6! is the same as the number of possible schedules.

c. 6! means the product of all integers between 1 and 6, which is exactly how we calculated the number of different schedules.

d. Because
$$
4!=4cdot 3 cdot 2cdot 1
$$

which is the same way we would calculate the number of possible ways to arrange the letters of MATH(since if you picked the first letter, there are three possibilities left for the second, then two possibilities left for the third and one possibility for the forth).

a. 40,320; 5,040; 720; 120; 24; 6; 2; 1

b. 6!

c. Product of all integers from 1 to 6.

d. $4!=4cdot 3cdot 2cdot 1$.

Using the definition of the factorial:

a.
$$
dfrac{10!}{8!}=10cdot 9 = 90
$$

b.
$$
dfrac{70!}{68!}=70cdot 69 = 4,830
$$

c.
$$
dfrac{7!}{4!3!}=dfrac{7cdot 6cdot 5}{3cdot 2 cdot 1} = 35
$$

d.
$$
dfrac{20!}{18!2!}=dfrac{20cdot 19}{2 cdot 1} = 190
$$

a. For the first letter there are 4 possibilities, the second letter 3 possibilities, the third letter 2 possibilities and the forth letter 1 possibility:

$$
4!=4cdot 3 cdot 2 cdot 1=24
$$

b.

c. NAPS, PANS, SNAP and SPAN are possible words. The probability is the nubmer of favorable outcomes divided by the number of possible outcomes:

$$
P(word)=dfrac{4}{24}=dfrac{1}{6}approx 0.167=16.7%
$$

a. 24 arrangements

b. Tree diagram

c. $frac{1}{6}approx 0.167=16.7%$

For the first speech there are 5 possibilities, the second speech 4 possibilities, the third speech 3 possibilities, etc.:

$$
5!=5cdot 4cdot 3 cdot 2 cdot 1=120
$$

$$
120
$$

a. The proportions of intersection chords are equal:

$$
dfrac{8}{k}=dfrac{k}{18}
$$

Use cross multiplication:

$$
144=k^2
$$

Take the square root of both sides of the equation and take into account that a negative length is not possible:

$$
12=k
$$

b. The total circle is 360$text{textdegree}$:

$$
z=360text{textdegree}-50text{textdegree}=310text{textdegree}
$$

The sine ratio is the opposite side divided by the hypotenuse:

$$
r=dfrac{5}{sin{25text{textdegree}}}approx 11.83
$$

a. $k=12$

b. $r=11.83$

The interior angle of a $n$-gon is
$$
dfrac{n-2}{n}180text{textdegree}
$$

Replace $n$ with 8:

$$
dfrac{n-2}{n}180text{textdegree}=dfrac{6}{8}180text{textdegree}=135text{textdegree}
$$

a. Since the circumference is $2pi r$, we know that the radius is:

$$
r=dfrac{40}{2pi}=dfrac{20}{pi}
$$

The surface area of the sphere is then:

$$
S=4pi r^2=4pi dfrac{20^2}{pi^2}=dfrac{1600}{pi}approx 509ft^2
$$

Determine the volume of the sphere:

$$
V=dfrac{4pi r^3}{3}=dfrac{4pi 20^3}{3pi^3}=dfrac{32000}{3pi^2}approx 1,081ft^3
$$

b. The circumference is then the product of the circumference of the original ball and the cubic root of 2(ratio of volumes):

$$
C=sqrt[3]{2}cdot 40approx 50ft
$$

a. Surface area 509 ft$^2$, Volume 1081 ft$^3$

b. 50 ft

Subtract the two equations:

$$
0=x^2-5x-6
$$

Factorize:

$$
0=(x-6)(x+1)
$$

Zero product property:

$$
x-6=0text{ or }x+1=0
$$

Solve each equation to $x$:

$$
x=6text{ or }x=-1
$$

Determine $y$:

$$
y=2x+8=2(6)+8=20
$$

$$
y=2x+8=2(-1)+8=6
$$

Thus the solutions are $(6,20)$ and $(-1,6)$.

$(6,20)$ and $(-1,6)$

c. Quadratic model obtained from technology:
$$
y=-0.08308x^2+14.18884x-578.89853
$$

Evaluate at $x=95$:

$$
y=-0.08308(95)^2+14.18884(95)-578.89853=19.24427
$$

Thus the attendance is about 19.24427 thousand or 19,244.

a. For the first digit there are 10 choices, the second 9 and the third 8 and the forth 7:

$$
10cdot 9 cdot 8 cdot 7=5,040
$$

b. The product of the integer 1 to 6 are missing (and thus 6!).

c. You can thus rewrite this as:

$$
dfrac{10!}{6!}=10cdot 9 cdot 8 cdot 7=5,040
$$

a. 5,040

b. Product of 1 to 6, 6!

c. $frac{10!}{6!}$

a. Repetition is not allowed and tus we need a permutation:

$$
dfrac{52!}{(52-4)!}=dfrac{52!}{48!}=52cdot 51cdot 50cdot 49=6,497,400
$$

b. Repetition is not allowed and tus we need a permutation:

$$
7!=5,040
$$

c. Repetition is not allowed and tus we need a permutation:

$$
dfrac{36!}{(36-3)!}=dfrac{36!}{3!}=36cdot 35cdot 34=42,840
$$

d. Repetition is not allowed and tus we need a permutation:

$$
35cdot 35 cdot 34=41,650
$$

e. Repetition is allowed:

$$
36cdot 36cdot 36=36^3=46,656
$$

a. ABC, ACB, BAC, BCA, CAB, CBA. The list is different, because in this list a letter cannot be repeated, while in the given list a letter can be repeated. This then means that the list of the child is a permutation.

b. The list of candidates will be longer than the list of applications, because the order matters with president/vice/secratary but it does not matter with the applications (group of applicant 1,2 and 3 is the same as the group of applicant 3,2 and 1).

c. The second situation is a permutation, because the tiles cannot be repeated.

d. Order is important and there cannot be repetition. In general, if you have to pick k from p:

$$
_pP_k=dfrac{p!}{(p-k)!}
$$

a. Letter cannot be repeated, List of the child is permutation

b. List of candidates will be longer

c. Second situation is permutation

d. $_pP_k=frac{p!}{(p-k)!}$

a. 52 is the number of contestants and 48 is the nubmer of contestants decreased by the number of contestants to pick.

b.
$$
_nP_r=dfrac{n!}{(n-r)!}
$$

c. (i)
$$
_7P_4=dfrac{7!}{(7-4)!}=7cdot 6 cdot 5cdot 4=840
$$

(ii)
$$
_{52}P_4=dfrac{52!}{(52-4)!}=52cdot 51 cdot 50cdot 49=6,497,400
$$

(iii)
$$
_{16}P_7=dfrac{16!}{(16-7)!}=16cdot 15 cdot 14cdot 13cdot 12cdot 11 cdot 10=57,657,600
$$

a. 52 is the number of contestants, 48 is the number of contestants that weren’t picked

b. $_nP_r=frac{n!}{(n-r)!}$

c. (i) 840 (ii) 6,497,400 (iii) 57,657,600

a. Letters cannot be repeated and order is important:

$$
_4P_4=dfrac{4!}{(4-4)!}=4!=24
$$

b. Since two of the letters are the same, every combination appears twice and thus there are only 12 combinations.

c. Since three letters are identical, there are only 4 possibilities (since A can be in the first to forth place while the remaining places are S).

d.
$$
(b) dfrac{4!}{(4-2)!}=dfrac{4!}{2!}
$$

$$
(c)dfrac{4!}{(4-1)!}=dfrac{4!}{3!}
$$

e. If a letter is repeated $k$ times, then divide the number of arrangements by $k!$ (as there are $k!$ ways to order $k$ elements).

a. 24 ways

b. 12 ways, two letters are the same

c. 4 ways

d. (b) $frac{4!}{2!}$ (c) $frac{4!}{3!}$

e. Divide the number of arrangements by $k!$

a. Order is important and letters cannot be repeated:
$$
_7P_7=dfrac{7!}{(7-7)!}=7!=5,040
$$

b. The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
dfrac{1}{5,040}approx 0.0002=0.02%
$$

a. Order is important and letters cannot be repeated:
$$
_{12}P_3=dfrac{12!}{(12-3)!}=dfrac{12!}{9!}=12cdot 11 cdot 10=1,320
$$

b. See (a)

Since $triangle ABD$ is a right triangle (definition tangent), we can use the Pythagorean theorem:

$$
DB=sqrt{AB^2+AD^2}=sqrt{4^2+3^2}=sqrt{25}=5
$$

The minimal distance from point $B$ to the hot tub is then $DB$ decreased by the radius:

$$
DB-r=5-3=2ft
$$

Determine the volume of the pyramid:

$$
V=dfrac{Bcdot h}{3}=dfrac{(12cdot 7)cdot 21}{2}=882 units^3
$$

a. Add and subtract 9:

$$
f(x)=(x^2+6x+9)+11-9
$$

Factorize the perfect square trinomial ($a^2pm 2ab+b^2=(apm b)^2$):

$$
f(x)=(x+3)^2+2
$$

b. The vertex is $(-3,2)$.

c. Since the graph does not intersect the $x$-axis, the equation has no real solutions.

a. $f(x)=(x+3)^2+2$

b. $(-3,2)$

c. No real solutions

a. ABCD is a square because the length of each side is:

$$
sqrt{4^2+2^2}=sqrt{20}=2sqrt{5}
$$

b. The rotation changes the sign of the $x$ and $y$ coordinates:

$$
C’=(-5,-8)
$$

The reflection then changes the sign of the $y$-coordinate:

$$
D”=(-7,4)
$$

a. Square

b. $C’=(-5,-8)$, $D”=(-7,4)$

a. The graph is the graph of $f(x)=x^2$ reflected about the $x$-axis:
$$
y=-f(x)=-x^2
$$

b. The graph is the graph of $f(x)=x^2$ translated right by 3 units:
$$
y=f(x-3)=(x-3)^2
$$

c. The graph is the graph of $f(x)=x^2$ translated down by 3 units:
$$
y=f(x)-3=x^2-3
$$

d. The graph is the graph of $f(x)=x^2$ stretched horizontally by factor $dfrac{1}{4}$:
$$
y=f(2x)=(2x)^2=4x^2
$$

a. Reflected about $x$-axis

b. Translated to the right by 3 units

c. Translated down by 3 units

d. Stretched horizontally by factor $frac{1}{4}$.

a. If $x$ is the time in months and $y$ the value of your investment. Then the function describes the value of your initial investment of $2,000 at a monthly interest rate of 4%.

\
\

b. If$x$is the number of years and$y$ the population of a town, then the function described that the initial population of the town is 25,000 and yearly the population decreases by 17%.

a. Order is important and repetition is not possible, thus we can use a permutation:

$$
_5P_3=dfrac{5!}{(5-3)!}=5cdot 4=20
$$

b. Using the initials of their names, we obtain the possibilities:

ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE

Thus we note that there are 10 possibilities.

c. (i)
$$
_6P_3=dfrac{6!}{(6-3)!}=6cdot 5cdot 4=120
$$

(ii) Using the initials of their names, we obtain the possibilities:

ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, DEF

Thus we note that there are 20 possibilities.

d. (i)
$$
_6P_4=dfrac{6!}{(6-4)!}=6cdot 5=30
$$

(ii)
$$
4!=4cdot 3 cdot 2 cdot 1 = 24
$$

a. The number of combinations is less than the number of permutations in each situation.

b. The permutation is divided by the factorial of the number of items you choose.

c. We would expect the combinations to be half the permutations, because every combination will appear twice in the list of permutations:

$$
_6P_2=dfrac{6!}{(6-2)!}=6cdot 5 cdot 4 cdot 3=360
$$

$$
_6C_2=dfrac{6!}{(6-2)!2!}=dfrac{360}{2}=180
$$

d.
$$
_nC_r=dfrac{n!}{(n-r)!r!}=frac{_nP_r}{r!}
$$

a. Number of combinations is less than number of permutations.

b. The permutation is divided by the factorial of the number of items you choose.

c. 360 ways, 180 ways, Yes

d. $_nC_r=dfrac{n!}{(n-r)!r!}=frac{_nP_r}{r!}$

a. Yes,

$$
_{20}C_4=dfrac{20!}{(20-4)!4!}=dfrac{20cdot 19cdot 18cdot 17}{4cdot 3cdot 2cdot 1}=4,845
$$

b. The formula has been derived from the formula of permutations, thus basically we first determine the number of permutations and then divided by $r!$ which eliminates the duplicate possibilities in the permutations.

a. Order is not important and repetition is not possible, thus we need to use a combination:

$$
_{56}C_6=dfrac{56!}{(56-6)!6!}=dfrac{56cdot 55cdot 54cdot 53cdot 52cdot 51}{6cdot 5cdot 4cdot 3cdot 2cdot 1}=32,468,436
$$

b. The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
P(win)=dfrac{1}{32,468,436}=0.00000003=0.000003%
$$

a. Order is not important and repetition is not possible, thus we need to use a combination:

$$
_{52}C_5=dfrac{52!}{(52-5)!5!}=dfrac{52cdot 51cdot 50cdot 49cdot 48}{ 5cdot 4cdot 3cdot 2cdot 1}=2,598,960
$$

b. Order is not important and repetition is not possible, thus we need to use a combination:

$$
_{13}C_5=dfrac{13!}{(13-5)!5!}=dfrac{13cdot 12cdot 11cdot 10cdot 9}{ 5cdot 4cdot 3cdot 2cdot 1}=1,287
$$

Since there are four different suits, the total number of possibilities is:
$$
4cdot 1,287=5,148
$$

Order is important and repetition is not possible for the choice of president and vice-president:

$$
_{10}P_2=dfrac{10!}{(10-2)!}=10cdot 9=90
$$

Order is not important and repetition is not possible, for the other three committee members:

$$
_{8}C_3=dfrac{8!}{(8-3)!3!}=dfrac{8cdot 7cdot 6}{ 3cdot 2cdot 1}=56
$$

$textbf{Fundamental counting principle}$: If the first event could occur in $m$ ways and the second event could occur in $n$ ways, then the number of ways that the two events could occur in sequence is $mcdot n$.

$$
_{10}P_2cdot _8C_3=90cdot 56=5040
$$

Thus there are 5040 ways.

a.
$$
0!=1
$$

b. $_8P_8$ means picking 8 items out of 8 and thus we should have:

$$
_8P_8=8!
$$

But since we also know:

$$
_8P_8=dfrac{8!}{(8-8)!}=dfrac{8!}{0!}
$$

We then need $0!=1$.

c. $dfrac{3!}{3}=2!$, $dfrac{2!}{2}=1!$, $dfrac{1!}{1}=0!$ and thus we then obtain that
$$
0!=dfrac{1!}{1}=dfrac{1}{1}=1
$$

a. 1

b. Picking 8 items out of 8, 8!

c. $0!=frac{1!}{1}$

A permutation is defined as:

$$
_nP_r=dfrac{n!}{(n-r)!}
$$

A combination is defined as:

$$
_nC_r=dfrac{n!}{(n-r)!r!}
$$

a.
$$
_{10}P_8=dfrac{10!}{(10-8)!}=10cdot 9cdot 8cdot 7cdot 6 cdot 5cdot 4cdot 3=1,814,400
$$

b.
$$
_{10}C_8=dfrac{10!}{(10-8)!8!}=dfrac{10cdot 9}{2cdot 1}=45
$$

c.
$$
_{6}C_1=dfrac{6!}{(6-1)!1!}=dfrac{6}{1}=6
$$

a. $1,814,400$

b. 45

c. 6

a. Determine the ratio of the weights:

$$
dfrac{700}{5.6}=125=5^3
$$

Determine the ratios of the surface area:

$$
dfrac{500}{20}=25=5^2
$$

Thus we note that the two nuggets are similar with linear scale factor 5.

b. The ratio of their lengths is equal to the linear scale factor and thus is 5.

a. Determine the discriminant:

$$
D=b^2-4ac=7^2-4(6)(-20)=49+480=529>0
$$

Since the discriminant is positive, the graph of the equation has two $x$-intercepts and two real roots.

b. Determine the discriminant:

$$
D=b^2-4ac=(-8)^2-4(1)(16)=64-64=0
$$

Since the discriminant is zero, the graph of the equation has one $x$-intercept and one real root.

c. Determine the discriminant:

$$
D=b^2-4ac=1^2-4(2)(3)=1-24=-23<0
$$

Since the discriminant is negative, the graph of the equation has zero $x$-intercepts and two complex roots.

d. Since the factorization of the function contains exactly one factor, the graph of the equation has one $x$-intercept and one real root.

a. Two $x$-intercepts, Two real roots.

b. One $x$-intercept, One real root.

c. Zero $x$-intercepts, Two complex roots

d. One $x$-intercept, One real root.

Determine the area of the base

A pentagon has 5 vertices and thus this pentagon has sides of 4mm. Divide the pentagon into 5 congruent triangles. Determine the height of these triangles using the tangent ratio:

$$
h=2tan{54text{textdegree}}approx 2.75
$$

The area of a triangles is the product of the base and the height divided by 2:

$$
dfrac{2.75cdot 4}{2}=5.5
$$

Since the pentagon is made up out of 5 triangles, the area of the pentagon is:

$$
5cdot 5.5=27.5mm^2
$$

The volume of a right pyramid is the product of the base and the height divided by 3:

$$
V=dfrac{Bcdot h}{3}=dfrac{27.5cdot 7}{3}approx 64mm^3
$$

Determine the surface area:

$$
S=27.5+5cdot dfrac{4cdot sqrt{3.4^2+7^2}}{2}approx 90 mm^2
$$

Volume 64 mm$^3$ and Surface area 90 mm$^2$

a. Both segments are the radius of the circle:

$$
5m+1=3m+9
$$

Subtract $3m$ from both sides of the equation:

$$
2m+1=9
$$

Subtract 1 from both sides of the equation:

$$
2m=8
$$

Divide both sides of the equation by 2:

$$
m=4
$$

b. The central angle is double the inscribed angle:

$$
3x-9=2(x+4)
$$

Use distributive property:

$$
3x-9=2x+8
$$

Subtract $2x$ from both sides of the equation:

$$
x-9=8
$$

Add 9 to both sides of the equation:

$$
x=17text{textdegree}
$$

c. Use the pythagorean theorem:

$$
p^2=(p-2)^2+6^2
$$

Rewrite:

$$
p^2=p^2-4p+4+36
$$

Subtract $p^2$ from both sides of the equation:

$$
0=-4p+40
$$

Add $4p$ to both sides of the equation:

$$
4p=40
$$

Divide both sides of the equation by 4:

$$
p=10
$$

d. The sum of all angles in a quadrilateral is 360$text{textdegree}$:

$$
2t+9+5t+8t-10+3t+1=360
$$

Combine like terms:

$$
18t=360
$$

Divide both sides of the equation by 18:

$$
t=20text{textdegree}
$$

a. $m=4$

b. $x=17text{textdegree}$

c. $p=10$

d. $t=20text{textdegree}$

A. The area of a circle is $pi$ multiplied by the squared radius.
$$
A=pi r^2=pi 5^2=25pi approx 79
$$

B. The area of a square is the square of the side lengths.
$$
A=s^2=9^2=81
$$

C. The area of a trapezium is the sum of the length of the bases, multiplied by the height, and divided by 2.
$$
A=dfrac{(8+10)9}{2}=81
$$

D. The area of a rhombus is the product of the side length and the height.
$$
A=9cdot 8=72
$$

Thus we note that D has the least area.

a. The graph has been tranlasted down by 3 units, thus then is $h=0$ and $k=-3$:

$$
y=f(x)-3
$$

b. The graph has been translated to the right by 2 units and then is $h=2$ and $k=0$:

$$
y=f(x-2)
$$

a. $y=f(x)-3$

b. $y=f(x-2)$

On Friday, repetition is allowed and order is important in cones:

$$
5cdot 5cdot 5=5^3=125
$$

On Friday, repetition is allowed and order is not important in dishes:

AAA, AAB, AAC, AAD, AAE, ABB, ABC, ABD, ABE, ACC, ACD, ACE, ADD, ADE, AEE, BBB, BBC, BBD, BBE, BCC, BCD, BCE, BDD, BDE, BEE, CCC, CCD, CCE, CDD, CDE, CEE, DDD, DDE, DEE, EEE

Thus we note that there are then 25 possibilities.

On Saturday, repetition is not allowed and order is important in cones:

$$
_5P_3=dfrac{5!}{3!}=5cdot 4=20
$$

On Saturday, repetition is not allowed and order is NOT important in dishes:

$$
_5C_3=dfrac{5!}{(5-3)!3!}=dfrac{5cdot 4}{2cdot 1}=10
$$

a. On Friday, repetition is allowed and order is important in cones, FUNDAMENTAL COUNTING PRINCIPLE:

$$
5cdot 5cdot 5=5^3=125
$$

On Saturday, repetition is not allowed and order is important in cones, PERMUTATION:

$$
_5P_3=dfrac{5!}{3!}=5cdot 4=20
$$

On Saturday, repetition is not allowed and order is NOT important in dishes, COMBINATION:

$$
_5C_3=dfrac{5!}{(5-3)!3!}=dfrac{5cdot 4}{2cdot 1}=10
$$

b. On Friday, repetition is allowed and order is not important in dishes:

AAA, AAB, AAC, AAD, AAE, ABB, ABC, ABD, ABE, ACC, ACD, ACE, ADD, ADE, AEE, BBB, BBC, BBD, BBE, BCC, BCD, BCE, BDD, BDE, BEE, CCC, CCD, CCE, CDD, CDE, CEE, DDD, DDE, DEE, EEE

Thus we note that there are then 25 possibilities.

$$
text{color{white} jmlirjtet mjretm lijretm lrejtj erjtl jrljtlr jlktjrl jtrlj ljr litelrj tlretm ermthmre hmtlrel thmrleht hlirht lhr}
$$

Repetition is not possible and order is important, thus we can use permutations:

$$
_7P_3=dfrac{7!}{(7-3)!}=7cdot 6cdot 5=210
$$

Order is not important and repetition is not allowed, thus we can use combinations:

$$
_8C_3=dfrac{8!}{(8-3)!3!}=dfrac{8cdot 7cdot 6}{3cdot 2cdot 1}=56
$$

$$
_{10}C_3=dfrac{10!}{(10-3)!3!}=dfrac{10cdot 9cdot 8}{3cdot 2cdot 1}=120
$$

$textbf{Fundamental counting principle}$: If the first event could occur in $m$ ways and the second event could occur in $n$ ways, then the number of ways that the two events could occur in sequence is $mcdot n$.

$$
begin{align*}
_8C_3cdot _{10}C_3&=56cdot 120=6720
end{align*}
$$

Thus there are $6720$ possibilities in total.

Order is important and repetition is not allowed, thus we can use permutation:

$$
_4P_1=dfrac{4!}{(4-1)!1!}=4
$$

$$
_7P_1=dfrac{7!}{(7-1)!1!}=7
$$

$$
_1P_1=dfrac{1!}{(1-1)!1!}=1
$$

$$
_2P_1=dfrac{2!}{(2-1)!1!}=2
$$

$$
_3P_1=dfrac{3!}{(3-1)!1!}=3
$$

$textbf{Fundamental counting principle}$: If the first event could occur in $m$ ways and the second event could occur in $n$ ways, then the number of ways that the two events could occur in sequence is $mcdot n$.

$$
begin{align*}
_4P_1cdot _7P_1cdot _1P_1cdot _2P_1cdot _3P_1&=4cdot 7cdot 1cdot 2cdot 3=168
end{align*}
$$

Thus there are $168$ possibilities in total.

a. Repetition is not allowed and order is important, thus we need to use a permutation:

$$
_{22}P_3=dfrac{22!}{(22-3)!}=22cdot 21cdot 20=9,240
$$

b. No, because a combination was not used since order is important.

c. Repetition is allowed and order is important, thus use the Fundamental Counting Principle:

$$
22cdot 22cdot 22=22^3=10,648
$$

d. Thus for the first number you would have 22 possibilities, the second number 21 possibilities and the last number also 21 possibilities (since the first number is possible again):

$$
22cdot 21cdot 21=9,702
$$

a. The sum of all angles in a triangle is 180$text{textdegree}$ and the inscribed angles on the same arc are congruent:

$$
x=180text{textdegree}-91text{textdegree}-12text{textdegree}=77text{textdegree}
$$

b. Determine the radius using the Pythagorean theorem:

$$
(18+BC)^2=BC^2+24^2
$$

Rewrite:

$$
324+36BC+BC^2=BC^2+576
$$

Subtract $BC^2$ from both sides of the equaton:

$$
36BC+324=576
$$

Subtract 324 from both sides of the equation:

$$
36BC=252
$$

Divide both sides of the equation by 36:

$$
BC=7
$$

The area of the circle is then:

$$
A=pi r^2=pi 7^2=49pi approx 154
$$

c. The proportions of intersecting chords is equal:

$$
dfrac{x}{6}=dfrac{10}{2x}
$$

Use cross multiplication:

$$
2x^2=60
$$

Divide both sides of the equation by 2:

$$
x^2=30
$$

Take the square root of both sides of the equation:

$$
x=pm sqrt{30}
$$

Only a positive length makes sense:

$$
x=sqrt{30}
$$

a. $x=77text{textdegree}$

b. $A=154$

c. $x=sqrt{30}$

a. The radius is the product of the radius of Earth and the scale factor:

$$
r=109cdot 4000=436,000mi
$$

b. The radius of the sun is greater than (and almost double) the distance between Earth and the Moon.

c. The ratio of the volume of the sun to the earth is the linear scale factor cubed:

$$
109^3=1,295,029
$$

Thus the earth fits about 1,295,029 times inside the Sun.

a. Rewrite the equation in graphing form. Add and subtract 9:

$$
f(x)=(x^2-6x+9)+5-9
$$

Factorize the perfect square trinomial ($a^2pm 2ab+b^2=(apm b)^2$):

$$
f(x)=(x-3)^2-4
$$

Thus the vertex is $(3,4)$.

b. The domain of a quadratic equation contains all real numbers $mathbb{R}$.

Since $a=1$ is positive, the graphs opens upwards and thus the vertex is a minimum. The range is then:

$$
[4,+infty)
$$

c. Since $a=1$ is positive, the graph opens upwards and thus the vertex is a minimum.

d. Determine the roots of the function:

$$
0=(x-3)^2-4
$$

Add 4 to both sides of the equation:

$$
4=(x-3)^2
$$

Take the square root of both sides of the equation:

$$
pm 2 =x-3
$$

Add 3 to both sides of the equation:

$$
5text{ or } 1=3pm 2=x
$$

The function is then positive on $(-infty, 1)$ and $(5,+infty)$.

a. $(3,4)$

b. Domain $mathbb{R}$, Range $[4,+infty)$

c. Minimum

d. $(-infty, 1)$ and $(5, +infty)$

a. Subtract the two equations:

$$
0=x^2-2x+5
$$

Determine the roots using the quadratic formula:

$$
x=dfrac{2pmsqrt{(-2)^2-4(1)(5)}}{2(1)}=1pm 2i
$$

Determine $y$:

$$
y=2x=2(1pm 2i)=2pm 4i
$$

Thus the solutions are $(1pm 2i, 2pm 4i)$.

a. $(1pm 2i, 2pm 4i)$

b. The graphs do not intersect.

a. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{45text{textdegree}}=dfrac{1}{1}=1
$$

b. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{45text{textdegree}}=dfrac{1}{sqrt{2}}=dfrac{sqrt{2}}{2}
$$

c. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{45text{textdegree}}=dfrac{1}{sqrt{2}}=dfrac{sqrt{2}}{2}
$$

a. 1

b. $frac{sqrt{2}}{2}$

c. $frac{sqrt{2}}{2}$

a) The problem asks us to find the probability that the sheep will get sick if she is free to move through the entire field compared to the probability that she will get sick if she is restrained.

b) The sheep’s grazing region is the entire field if she is free, and her grazing region is a circle of radius of $20$ ft if she is restrained as described in the exercise.

c) In order to calculate the probability for both cases, we need to calculate the following:
$1.$ The number of bites she eats in the given time.
$2.$ The area of the field available to the sheep.
$3.$ The number of possible bites in the available area.
$4.$ The ratio between the number of bites the sheep will eat in the given period of time and the total number of bites available (this is the probability we need to calculate).

Based on the diagram, the area of the region that Zoe can roam at point M.
$$begin{aligned}
A_M&=dfrac{270degree}{360degree}pi r^2\
A_M&=0.75(3.1416)(20)^2\
A_M&=942.48medspacetext{ft}^2
end{aligned}$$

Take note that area $A_p$ is only one quarter of the circle with radius of $10medspacetext{ft}$ within the $10times10$ square as shown in diagram in **Step 1**. Therefore, the total area of region that Zoe can roam is.
$$begin{aligned}
A_R&=A_M+A_P\
A_R&=942.48+left(dfrac{90degree}{360degree}cdotpicdot(10)^2right)\
A_R&=942.48+78.54\
A_R&=1021.02medspacetext{ft}^2
end{aligned}$$

**b**. Since the given is $40$ bites per square foot, the total bites of plant within the region Zoe can reach is.
$$begin{aligned}
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot A_R\
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot 1021.02medspacetext{ft}^2\
text{Total bites}&approx40841medspacetext{bites}
end{aligned}$$

**c**. Since Zoe takes about one bite of plants every $3$ minutes for six hours a day. The total bites of food that Zoe eat each day is.
$$begin{aligned}
text{bites per day}&=dfrac{1medspacetext{bite}}{3medspacetext{mins}}cdotdfrac{60medspacetext{mins}}{1medspacetext{hr}}cdotdfrac{6medspacetext{hr}}{text{day}}\
text{bites per day}&=120dfrac{text{bites}}{text{day}}
end{aligned}$$

**d**. Since the locoweed is within the region of Zoe can roam, the probability that Zoe eats the locoweed in one day is.
$$begin{aligned}
P_W&=dfrac{text{bites per day}}{text{total bites in area of Zoe can roam}}\
end{aligned}$$

where

$P_W$ – probability of Zoe can eat locoweed in one day.

Solving for the value of $P_W$.
$$begin{aligned}
P_W&=dfrac{120}{40841}\
P_W&=0.00294
end{aligned}$$

To find the probability of Zoe can eat the locoweed in the proposed region, we need to find the area of region that Zoe can roam.
$$begin{aligned}
A_R&=dfrac{270degree}{360degree}cdotpi r^2
end{aligned}$$

where

$A_R$ – area of the region that Zoe can roam.

Solving for the value of $A_R$.
$$begin{aligned}
A_R&=0.75cdot(3.1416)(10)^2\
A_R&=235.62medspacetext{ft}^2
end{aligned}$$

Since there are $40$ bites per square foot, the total bites in Zoe’s region is.
$$begin{aligned}
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot A_R\
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot235.62medspacetext{ft}^2\
text{Total bites}&approx9425medspacetext{bites}
end{aligned}$$

Since Zoe takes about one bite of plants every 3 minutes for six hours a day. The total bites of food that Zoe eat each day is.
$$begin{aligned}
text{bites per day}&=dfrac{1medspacetext{bite}}{3medspacetext{mins}}cdotdfrac{60medspacetext{mins}}{1medspacetext{hr}}cdotdfrac{6medspacetext{hr}}{text{day}}\
text{bites per day}&=120dfrac{text{bites}}{text{day}}
end{aligned}$$

Solving for the probability that Zoe can eat locoweed in one day.
$$begin{aligned}
P&=dfrac{text{bites per day}}{text{total bites}}\
P&=dfrac{120}{9425}\
P&=0.0127
end{aligned}$$

To solve the area of region that Zoe can roam, we calculate the area of right triangle $triangle{BXY}$. Solving for the value of $y$ in the diagram using Pythagorean Theorem.
$$begin{aligned}
r^2&=(20)^2+y^2\
y^2&=r^2-(20)^2\
y^2&=(30)^2-(20)^2\
y^2&=900-400\
y^2&=500
end{aligned}$$

Square root both sides of the equation.
$$begin{aligned}
sqrt{y^2}&=sqrt{500}\
y&=22.36medspacetext{ft}
end{aligned}$$

The area of triangle $triangle{BXY}$ is.
$$begin{aligned}
A_{triangle{BXY}}&=dfrac{1}{2}y(20)\
A_{triangle{BXY}}&=dfrac{1}{2}(22.36)(20)\
A_{triangle{BXY}}&=223.6medspacetext{ft}
end{aligned}$$

To get the area of sector of circle, we need to find the angle $angle{BA}$. Based on the diagram in **Step 7**, the formula for angle $angle{BA}$ can be written as.
$$begin{aligned}
angle{BA}&=90degree-angle{BX}
end{aligned}$$

Solving for the value of $angle{BX}$ using trigonometric ratio.
$$begin{aligned}
sinangle{BX}&=dfrac{y}{r}\
sinangle{BX}&=dfrac{22.36}{30}\
sinangle{BX}&=0.745\
angle{BX}&=arcsin0.745\
angle{BX}&=48.16degree
end{aligned}$$

Solving for the value of $angle{BA}$.
$$begin{aligned}
angle{BA}&=90degree-angle{BX}\
angle{BA}&=90degree-48.16degree\
angle{BA}&=41.84degree
end{aligned}$$

Solving for the area of region that Zoe can roam.
$$begin{aligned}
A_R&=A_{triangle{BXY}}+Atext{(sector YBA)}\
A_R&=223.6+dfrac{angle{BA}}{360degree}pi r^2\
A_R&=223.6+dfrac{41.84degree}{360degree}cdot(3.1416)(30)^2\
A_R&=223.6+0.116(2827.44)\
A_R&=223.6+327.98\
A_R&=551.58medspacetext{ft}^2
end{aligned}$$

Since there are $40$ bites per square foot, the total bites in Zoe’s region is.
$$begin{aligned}
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot A_R\
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot551.58medspacetext{ft}^2\
text{Total bites}&approx22063medspacetext{bites}
end{aligned}$$

Solving for the probability that Zoe can eat locoweed in one day.
$$begin{aligned}
P&=dfrac{text{bites per day}}{text{total bites}}\
P&=dfrac{120}{22063}\
P&=0.00544
end{aligned}$$

Based on the diagram, we need to solve first the area of $triangle{CEF}$, **area of sector** $CZG$ and $A_F$. Solving for the value of $A_{triangle{CEF}}$.
$$begin{aligned}
A_{triangle{CEF}}&=dfrac{1}{2}(10)(20)\
A_{triangle{CEF}}&=100medspacetext{ft}^2
end{aligned}$$

To solve for the area of sector $CZG$, we need to solve first the angle $angle{CF}$. Using trigonometric ratio to find the value of $angle{CF}$.
$$begin{aligned}
tantheta&=dfrac{20}{10}\
tantheta&=2\
theta&=arctan2\
theta&=63.43degree
end{aligned}$$

Solving for the length of $overline{CF}$ using Pythagorean Theorem.
$$begin{aligned}
overline{CF}&=sqrt{20^2+10^2}\
overline{CF}&=sqrt{400+100}\
overline{CF}&=sqrt{500}\
overline{CF}&=22.36medspacetext{ft}
end{aligned}$$

Solving for the length of $overline{FG}$.
$$begin{aligned}
overline{FG}&=r-overline{CF}\
overline{FG}&=30-22.36\
overline{FG}&=7.64medspacetext{ft}
end{aligned}$$

Solving for the area of region that Zoe can roam.
$$begin{aligned}
A_R&=A_F+A_{triangle{CEF}}+A_{text{sector}GS}+(A_{text{sector}ST}-A_{ABTC})\
A_R&=left(dfrac{theta}{360degree}cdotpi overline{FG}^2right)+100+left(dfrac{270degree-theta}{360degree}cdotpi r^2right)+left[dfrac{90degree}{360degree}cdotpi r^2-(30)(10)right]\
A_R&=left(dfrac{63.43degree}{360degree}cdot pi (7.64)^2right)+100+left(dfrac{270degree-63.43degree}{360degree}cdotpi (30)^2right)+left[dfrac{90degree}{360degree}cdot pi (30)^2-(30)(10)right]\
A_R&=32.31+100+1622.29+left[706.86-(300)right]\
A_R&=1754.6+406.86\
A_R&=2161.46medspacetext{ft}^2
end{aligned}$$

Since there are $40$ bites per square foot, the total bites in Zoe’s region is.
$$begin{aligned}
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot A_R\
text{Total bites}&=40dfrac{text{bites}}{text{ft}^2}cdot2161.46medspacetext{ft}^2\
text{Total bites}&approx86458medspacetext{bites}
end{aligned}$$

Solving for the probability that Zoe can eat locoweed in one day.
$$begin{aligned}
P&=dfrac{text{bites per day}}{text{total bites}}\
P&=dfrac{120}{86458}\
P&=0.00139
end{aligned}$$

**Given**
There are $5$ French books and $3$ Spanish books to be placed in a shelf, given that each language must remain together. Determine the number of ways these books can be arranged.

**Solution**
We first determine the number of ways we can arrange the five French books using permutations. There are $5$ French books so we start with $5$ choices for the first slot, then $4$ choices on the second slot, and so on. We obtain the number of ways we can place the French books as follows:
$$begin{aligned}
5 times 4 times 3 times 2 times 1 = 120 text{ ways}
end{aligned}$$

This expression can also be represented as a factorial form: $5!$

Next, we determine the number of ways we can arrange the Spanish books given the same method we did for the French books:
$$begin{aligned}
3 times 2 times 1 = 6 text{ ways}
end{aligned}$$

This expression can also be represented as a factorial form: $3!$

Next, we consider that each language must remain together, in this scenario, there are only two choice of placements, hence, we have the value $2!$

Lastly, to obtain the number of ways given the scenario in the problem, we multiply each value obtained from each criteria as follows:
$$begin{aligned}
&= 5! times 3! times 2! \
&= boxed{1440 text{ ways}}
end{aligned}$$

There are $1440 text{ ways}$ to arrange the books

$1440 text{ ways}$

**Concept**
Francis and John’s distance can be described by the the following equation:
$$begin{gather}
x_f = x_o + at
end{gather}$$

where $x_f$ is the distance traveled after time $t$, $x_o$ is the initial distance in front of the starting line, and $a$ is the running speed

**Given**
$x_{o,F} = 2 text{ m}$
$x_{o,J} =5 text{ m}$
$a_{F} = 1 frac{text{m}}{text{s}}$
$a_{J} = 0.75 frac{text{m}}{text{s}}$

Determine the time, $t$ when Francis will catch up to John

**Solution**
We can determine the time when Francis will catch up to John by finding the value of $t$ such that the final distance covered by the two are equal. We determine the expression for this scenario using Eq (1) as follows:
$$begin{aligned}
x_{f,F} &= x_{o,F} + a_{F} t \
x_{f,J} &= x_{o,J} + a_J t \
x_{o,F} + a_{F} t &= x_{o,J} + a_J t
end{aligned}$$

We plug-in the values to the expression then solve for the value of $t$ as follows:
$$begin{aligned}
x_{o,F} + a_{F} t &= x_{o,J} + a_J t \
2 + 1t &= 5 + 0.75t \
2 + t &= 5 + 0.75t \
t – 0.75t &= 5 – 2 \
0.25t &= 3 \
t &= frac{3}{0.25} \
&= boxed{12 text{ s}}
end{aligned}$$

Francis will be able to catch up to John after $12 text{ s}$

$t=12 text{ s}$

**Given**
We are given with the following equation that describes the scenario:
$$begin{gather}
y = -16x^2 + 64x + 80
end{gather}$$

where $y$ is the height travelled by the ball in feet, and $x$ is the time travelled in seconds

**Part a**
We determine how how is the ball when thrown from the edge of the cliff. We determine the value of the height when $x=0$ using Eq (1) as follows:
$$begin{aligned}
y &= -16x^2 + 64x + 80 \
y &= -16(0)^2 + 64(0) + 80 \
&= boxed{80 text{ feet}}
end{aligned}$$

This means that the ball is thrown initially from the edge of the cliff at a height of $80 text{ feet}$

**Part b**
Next, we obtain the height of the ball after $x=3 text{ s}$ and $x=0.5 text{ s}$ after it is thrown using Eq (1) as follows:
$$begin{aligned}
y_3 &= -16x^2 + 64x + 80 \
y_3 &= -16(3)^2 + 64(3) + 80 \
&= -144 +192 +80 \
&= boxed{128 text{ feet}}
end{aligned}$$

After $3 text{ s}$, the ball is at a height $128 text{ feet}$

$$begin{aligned}
y_{0.5} &= -16x^2 + 64x + 80 \
y_{0.5} &= -16(0.5)^2 + 64(0.5) + 80 \
&= -4 +32 +80 \
&= boxed{108 text{ feet}}
end{aligned}$$

After $0.5 text{ s}$, the ball is at a height $108 text{ feet}$

**Part c**
When the ball hits the ground, the height of the ball will be equal to $0 text{ ft}$. We determine the time, $x$ when this happens by setting $y=0$. We solve for the value of $x$ for this scenario using Eq (1) as follows:
$$begin{aligned}
y &= -16x^2 + 64x+ 80 \
0 &= -16x^2 + 64x+ 80 \
0 &= frac{-16x^2}{-16} + frac{64x}{-16} + frac{80}{-16} \
0 &= x^2 -4x +5 \
0 &= (x-5)(x+1) \
x &= {-1,5 }
end{aligned}$$

Based from the results, the ball hits the ground at the time, $x = {-1,5 }$, we know that having a negative time for the scenario cannot happen, hence, we consider the positive value of $x$. Hence, the ball will hit the ground when $boxed{x=5}$

**Part d**
The domain of the function is $[0,5]$, this means that the possible values for the time ranges from $x=0$ to $x=5$. Any value does it not included in this range is not considered a solution of the equation or does not represent the scenario.

The two angles $angle{a}$ and $angle{b}$ are connected to same points in the circle which is point $B$ and $C$. When a polygon is inscribe in a circle, the formula for angles $a$ is.
$$begin{aligned}
angle{a}&=dfrac{1}{2}(overset{frown}{BC})\
end{aligned}$$

SInce $angle{a}$ will be smaller compare to the central angle which is equal to the $overset{frown}{BC}$, the $angle{b}$ which can be a central angle or not is bigger than $angle{a}$. Therefore,
$$begin{aligned}
(angle{a})<(angle{b})
end{aligned}$$

**b**. To determine which side is bigger, we need to solve the remaining angle. The sum of interior angles of a triangle is $180degree$. Therefore, the equation will be,

Let $x$ = remaining angle
$$begin{aligned}
62degree+70degree+x&=180degree\
x&=180degree-62degree-70degreemedspacetext{(Transposition Method)}\
x&=48degree
end{aligned}$$

Since $angle{x}$ which is opposite to side $b$ is smaller compare to the angle opposite side $a$ which has angle of $62degree$, side $a$ is greater than the side $b$.
$$begin{aligned}
a>b
end{aligned}$$

**c**. Since the triangle is an equilateral triangle because all sides are congruent, the formula for it’s area $a$ is,
$$begin{aligned}
a&=dfrac{sqrt{3}}{4}cdot6^2\
a&=15.59
end{aligned}$$

Since the polygon is a square, all sides are equal. To solve for it’s area $b$, the formula is,
$$begin{aligned}
b&=4^2\
b&=16
end{aligned}$$

Therefore, $a$<$b$.

The objective of this problem is to find the coordinate of the vertex point $C$ of an equilateral triangle and provide the solution in detail.

Since the two coordinates are given, the $x$ coordinate of vertex point $C$ can be solve by finding the median of two vertex points $A$ and $B$.

Given: $A$ ($0,0$), $B$($12,0$)
$$begin{aligned}
x&=dfrac{12-0}{2}\
x&=6
end{aligned}$$

To find the $y$ coordinate of vertex $C$, solve for the height $h$ of the triangle. Since the triangle is an equilateral triangle, the formula for height is,
$$begin{aligned}
h=dfrac{sqrt{3}}{2}a
end{aligned}$$

where $a$ is the length of one side of the triangle.

Since the $x$ coordinates of two vertices $A$ and $B$ are given which is $0$ and $12$ respectively, the length of one side is,
$$begin{aligned}
a&=x_B-x_A\
a&=12-0\
a&=12
end{aligned}$$

Solving for the $y$ coordinate of vertex $C$ by finding the height.
$$begin{aligned}
y&=dfrac{sqrt{3}}{2}a\
y&=dfrac{sqrt{3}}{2}(12)\
y&=6sqrt{3}
end{aligned}$$

See illustration below for the coordinates of vertex $C$.

**Concept**
The length of the sides $30^circ-60^circ-90^circ$ triangle or right triangle can be described using the Pythagorean theorem:
$$begin{gather}
c^2 = a^2 + b^2
end{gather}$$

**Given**
$c=2 text{ units}$

Sketch and label the triangle and determine the values of the following: (a.) $tan 30^circ$, (b.) $sin 30^circ$, (c.) $cos 30^circ$, and determine if the relationship of $sin 60^circ$ and $cos 30^circ$ is the same for any pair of acute angles in a right triangle

**Solution**

We assume that one side of the triangle has a value of $1 text{ unit}$. From the Pythagorean theorem we determine the value of the sides as follows:
$$begin{aligned}
c^2 = a^2 + b^2 \
2^2 &= 1^2 + b^2 \
4 &= 1 + b^2 \
4-1 &= b^2 \
b^2 &= 3 \
b &= sqrt{3} \
end{aligned}$$

The length of each side of the right triangle are the following values: $c=2$, $a=1$, and $b=sqrt{3}$

**Part a**
We obtain the value of $tan 30^circ$. We know that the trigonometric identity for $tan$ is $tan theta= frac{text{opposite}}{text{adjacent}}$. Hence, by using the length of the opposite and adjacent sides we can determine the value of $tan theta$.

We verify the value of $tan 30^circ$ using this property as follows:
$$begin{aligned}
tan 30^circ &= boxed{0.58} \
tan 30^circ &= frac{text{opposite}}{text{adjacent}} \
&= frac{1}{sqrt{3}} \
&= boxed{0.58}
end{aligned}$$

**Part b**
Next, we obtain the value of $sin 30^circ$. We know that the trigonometric identity for $sin$ is $sin theta= frac{text{opposite}}{text{hypotenuse}}$. Hence, by using the length of the opposite and adjacent sides we can determine the value of $sin theta$.

We verify the value of $sin 30^circ$ using this property as follows:
$$begin{aligned}
sin 30^circ &= frac{1}{2} = boxed{0.5} \
sin 30^circ &= frac{text{opposite}}{text{hypotenuse}} \
&= frac{1}{2} \
&= boxed{0.5}
end{aligned}$$

**Part c**
Next, we obtain the value of $cos 30^circ$. We know that the trigonometric identity for $cos$ is $cos theta= frac{text{adjacent}}{text{hypotenuse}}$. Hence, by using the length of the opposite and adjacent sides we can determine the value of $cos theta$.

We verify the value of $cos 30^circ$ using this property as follows:
$$begin{aligned}
cos 30^circ &= frac{sqrt{3}}{2} = boxed{0.87} \
cos 30^circ &= frac{text{adjacent}}{text{hypotenuse}} \
&= frac{sqrt{3}}{2} \
&= boxed{0.87}
end{aligned}$$

**Part d**
We obtain the value of $sin 60^circ$ and $cos 30^circ$ from the diagram as follows:
$$begin{aligned}
sin 60^circ &= frac{text{opposite}}{text{hypotenuse}} \
&= frac{sqrt{3}}{2} \
cos 30^circ &= frac{text{adjacent}}{text{hypotenuse}} \
&= frac{sqrt{3}}{2} \
end{aligned}$$

We observe the the two have the same values.

We determine if this relationship is true for any pair of acute angles in the right triangle, we obtain the value of $sin 30^circ$ and $cos 60^circ$ from the diagram as follows:
$$begin{aligned}
sin 30^circ &= frac{text{opposite}}{text{hypotenuse}} \
&= frac{1}{2} \
cos 60^circ &= frac{text{adjacent}}{text{hypotenuse}} \
&= frac{1}{2} \
end{aligned}$$

We observe the the two have the same values. Hence, this relationship holds true for any pair of acute angles in a right triangle.

**Concept**
The daily compounded interest can be described by the equation:
$$begin{gather}
f(t) = a times left(1+frac{r}{365} right)^{tcdot 365}
end{gather}$$

**Given**
First Equations Bank: $f(t)= a(1.04)^t$
You Figure Bank: $r = 3.9%$

Determine what type of account will Sofia choose

**Solution**
We can determine which bank should Sofia choose by getting which bank produced the highest interest gained from her deposit. We assume that Sofia invested an initial amount of $$ 10000$, we determine the amount after $t=5 text{ yrs}$ from First Equations Bank as follows:
$$begin{aligned}
f(t)&= a(1.04)^t \
f(5)&= 10000 cdot (1.04)^5 \
&= boxed{12166.53}
end{aligned}$$

Next, we obtain the amount of Sofia’s investment from the You Figure Bank using Eq (1) as follows:
$$begin{aligned}
f(t) &= a times left(1+frac{r}{365} right)^{tcdot 365} \
f(t) &= 10000 times left(1+frac{0.039}{365} right)^{5cdot 365} \
&= boxed{12152.98}
end{aligned}$$

Based from the results, the money invested in First Equations Bank after a period of $5 text{ yrs}$ is greater than the money invested in You Figure Bank. Hence, Sofia should choose to invest in First Equations Bank.

By using the formula of ratio of similarities for volume.

Let
$x$ = new volume
$y$ = original volume
$$begin{aligned}
x=r^3cdot y
end{aligned}$$

where $r$ is linear scale factor.

Solving for $r$.
$$begin{aligned}
702&=r^3cdot 26
end{aligned}$$

Divide both sides by $26$.
$$begin{aligned}
dfrac{702}{26}&=dfrac{r^3cdot 26}{26}\
27&=r^3
end{aligned}$$

Taking cube root for both sides of the equation.
$$begin{aligned}
sqrt[3]{27}&=sqrt[3]{r^3}\
3&=r
end{aligned}$$

**Solution**
We use permutations to determine the number of ways Mike can arrange the ten layers. We consider that in the first slot, there are $10$ choices, in the second slot, there are $9$ choices and so on. The number of ways for the layers to be arranged is as follows:
$$begin{aligned}
& 10 times 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 = boxed{3628800 text{ ways}}
end{aligned}$$

This expression can be expressed as $10!$

We take note that for Mike to build the ten-layer dip, there should be layers that are arranged in such a way that they are together. He has the three layers of mashed avocado and two layers of cheddar cheese. The number of ways to arrange the mashed avocado is:
$$begin{aligned}
3 times 2 times 1 = boxed{6 text{ ways}}
end{aligned}$$

As for the two layers of cheese we can arranged these $2$ ways. We divide the number of ways Mike can arrange the ten-layer with the considerations. The number of ways Mike can arrange the ten-layer dip with the considerations can be obtained as follows:
$$begin{aligned}
& frac{10!}{3! 2} = boxed{302400 text{ ways}}
end{aligned}$$

$302400 text{ ways}$

**a**. The equation given is an absolute value function and the formula is,
$$begin{aligned}
y=a|x-h|+k
end{aligned}$$
where $(h, k)$ is the vertex of the graph.

The given equation is,
$$begin{aligned}
y=|x-1|enspacetext{(where $a$ = 1, $h$ = 1 and $k$ = 0)}
end{aligned}$$
The vertex of the graph is $(1,0)$.

**b**. Using the same steps, the vertex of the given equation is,
$$begin{aligned}
y=|x|+3enspacetext{($a$ = 1, $h$ = 0 and $k$ = 3)}
end{aligned}$$
The vertex of the graph is $(0,3)$.

**c**. Using the same steps, the vertex of the given equation is,
$$begin{aligned}
y=|x-1|+3enspacetext{($a$ = 1, $h$ = 1 and $k$ = 3)}
end{aligned}$$
The vertex of the graph is $(1,3)$.

**Concept**
The volume of a rectangular prism can be obtained using:
$$begin{gather}
V_{prism} = lwh
end{gather}$$

On the other hand, the volume of a cylinder can be calculated using:
$$begin{gather}
V_{cylinder} = pi r^2 h
end{gather}$$

**Part a**
We determine the volume of the solid by getting the volume of the prism and subtracting the volume of the hole (cylinder). We first obtain for the expression for the volume of the solid using Eq (1) and Eq (2) as follows:
$$begin{aligned}
V_{solid} &= V_{prism} – V_{cylinder} \
V_{solid} &= lwh – (pi r^2 h) \
end{aligned}$$

From the figure, we obtain the value of the height, width, length, and radius. Using the expression, we obtain the volume of the solid as follows:
$$begin{aligned}
V_{solid} &= l w h – (pi r^2 h) \
&= (5) cdot (2) cdot (6) – (pi (0.5)^2 (6)) \
&= 60 – (4.71) \
&= boxed{55.28 text{ cm}^3}
end{aligned}$$

The volume of the solid is $55.28 text{ cm}^3$

**A**
First observe the upper triangle, which has an angle of $80°$. Properly label its sides and vertices, like shown in the figure.

Use the Law of Cosines to verify if the length of the side $a$ really is $5$ given the measures of the other sides and the angle $A$.
$$begin{aligned}a=5, b=7, c= 4, A=80°end{aligned}$$
$$begin{aligned}
a^2&=b^2+c^2-2bccos A\
5^2&=7^2+4^2-2cdot7cdot4cos(80°)\
25&=49+16-56cdot0.173648\
25&=65-9.72\
25&ne55.27
end{aligned}$$
This is proof that the triangle in the figure cannot have the given measures.

Repeat the procedure for the other given triangle.

$$begin{aligned}a=5, b=4, c= 7, A=80°end{aligned}$$
$$begin{aligned}
a^2&=b^2+c^2-2bccos A\
5^2&=4^2+7^2-2cdot4cdot7cos(50°)\
25&=16+49-56cdot0.64279\
25&=65-25.996\
25&ne29.0
end{aligned}$$

**B**

Look at the given figure. It is made of two triangles.

Calculate the value of the angle $x$ if you know that the sum of the angles in the triangle equals $180°$.
$$begin{aligned}
x+40°+50°&=180°\
x&=180°-40°-50°\
x&=90°
end{aligned}$$
The angle symmetrical to the angle $x$ should also have a measure of $90°$, it cannot possibly have a measure of $80°$.

**C**
The given is an isosceles triangle since the line from the upper vertex passing through the bottom side of the triangle bisects the bottom side into two equal halves. We know that an isosceles triangle has both its arms equal. Therefore, one of the arms cannot possibly have a length of $10$, while the other arm has a length of $12$.

**D**
The given is an isosceles triangle since the arms of the triangle have the same length. We know that the angles the arms of an isosceles triangle enclose with the third side of the triangle must have an equal measure. Therefore, in no way can one of these angles have a measure of $50°$ while the other angle has a measure of $60°$.

**E**
We are given a geometric figure comprised of two identical right triangles. We know that because each of these right triangles has a right angle and they have another identical angle (labeled with the lines). Since we know that the sum of the angles in any triangle is $180°$, if two angles have equal measures for two angles, they also must have equal measures of the third triangle. Therefore, it is impossible that one right triangle has a leg with a measure of $13$, while the same leg of the other triangle has a measure of $14$ because the measures of the angles would not be the same.

**Solution**
Solve the equation by completing the square.
$$begin{aligned}
x^2-6x+11 &= 0 \
x^2-6x+11-11 &= 0-11 \
x^2-6x+ left(frac{6}{2} right)^2 &= -11 \
x^2-6x+ 3^2 &= -11+3^2 \
x^2-6x+9 &= -11+9 \
(x-3)^2 &= -2 \
end{aligned}$$

Get the square root of both sides.
$$begin{aligned}
(x-3)^2 &= -2 \
sqrt{(x-3)^2} &= sqrt{-2} \
x-3 &= pm sqrt {-2} \
x-3+3 &= pm sqrt {-2}+3 \
x &= 3 pm sqrt {-2} \
end{aligned}$$

$x = 3 pm sqrt {-2}$

**Solution**
Compute for the area of the Moon.
$$begin{aligned}
A &= pi r^2 \
&= pi (1080)^2 \
&= pi cdot 1166400 \
&= 3664353.67 \
end{aligned}$$

Compute for the ratio of the area of the Moon to **Asia**.
$$begin{aligned}
text{Ratio} &= frac{text{Area of the Moon}}{text{Area of Asia}}\\
&= frac{17212048.1}{3664353.6} \\
&= boxed{4.69}
end{aligned}$$

Compute for the ratio of the area of the Moon to **Africa**.
$$begin{aligned}
text{Ratio} &= frac{text{Area of the Moon}}{text{Area of Africa}}\\
&= frac{11608161.4}{3664353.6} \\
&= boxed{3.17}
end{aligned}$$

Compute for the ratio of the area of the Moon to **North America**.
$$begin{aligned}
text{Ratio} &= frac{text{Area of the Moon}}{text{Area of North America}}\\
&= frac{9365294.0}{3664353.6} \\
&= boxed{2.56}
end{aligned}$$

Compute for the ratio of the area of the Moon to **South America**.
$$begin{aligned}
text{Ratio} &= frac{text{Area of the Moon}}{text{Area of South America}}\\
&= frac{6879954.4}{3664353.6} \\
&= boxed{1.88}
end{aligned}$$

Compute for the ratio of the area of the Moon to **Antarctica**.
$$begin{aligned}
text{Ratio} &= frac{text{Area of the Moon}}{text{Area of Antarctica}}\\
&= frac{5100023.4}{3664353.6} \\
&= boxed{1.39}
end{aligned}$$

Compute for the ratio of the area of the Moon to **Europe**.
$$begin{aligned}
text{Ratio} &= frac{text{Area of the Moon}}{text{Area of Europe}}\\
&= frac{3837083.3}{3664353.6} \\
&= boxed{1.05}
end{aligned}$$

Compute for the ratio of the area of the Moon to **Australia/Oceania**.
$$begin{aligned}
text{Ratio} &= frac{text{Area of the Moon}}{text{Area of Australia/Oceania}}\\
&= frac{2967967.3}{3664353.6} \\
&= boxed{0.81}
end{aligned}$$

**Solution**
Solve for $x$ by substituting $y$.
$$begin{aligned}
x^2 + y^2 &= 9 \
x^2 + (2)^2 &= 9 \
x^2 + 4 &= 9 \
x^2 + 4 -4 &= 9-4 \
sqrt{x^2} &= sqrt{5} \
x &= pm sqrt{5}\
end{aligned}$$

Notice that the first equation is a circle with center at the origin.
$$begin{aligned}
x^2+y^2 &= r^2 \
end{aligned}$$
becomes
$$begin{aligned}
x^2+y^2 &= 9 \
x^2+y^2 &= 3^2 \
end{aligned}$$
The circle is centered at $(0,0)$ and has radius $3$.

The second equation is a horizontal line at $y=2$.

With the details you know, you can conclude that a horizontal line intersects the circle at two points.
$$begin{aligned}
(sqrt{5},2) text{ and } (-sqrt{5}, 2)
end{aligned}$$

Use the graphing calculator to determine the positions of the relevant points.

Let’s assume that the house is located at the point $H(0,0)$ and the school and the practice field at the point $S(2,0)$.

Also, if the person is located three miles from the school, his distance from the school can be described by an equation of a circle with the origin at point $S$ with the radius of $3$, or
$$begin{aligned}
(x-p)^2+(y-q)^2&=r^2\
(x-2)^2+(y-0)^2&=3^2\
(x-2)^2+y^2&=9
end{aligned}$$


We know that the person is moving along the straight line, represented by the x-axis. Therefore, $y=0$ because the person never moves from the x-axis. Calculate the values of the variable $x$ from the above equation.
$$begin{aligned}
(x-2)^2+y^2&=9\
(x-2)^2+0&=9\
(x-2)^2&=9\
x-2&=pm3 quad*\
x&=2pm3\
x_1&=boxed{5}\
x_2&=boxed{-1}
end{aligned}$$
$*$Take square root of both sides.
Therefore, the person is located either on the point $A(-1,0)$ or the point $B(5,0)$ at the moment of the call.

These solutions could also be obtained graphically. Note that if at the moment of the phone call the person is located $3$ miles from the football practice, this is represented by the intersection of the circle and the x-axis on the above figure. Therefore, the person is located either at the point $A(-1,0)$ or the point $B(5,0)$ at the moment of the phone call, as indicated by the figure below.

Conclude that the person is located either one mile or five miles from his home at the moment of the phone call.

**Part b**
A linear model will best fit the data.
[](https://postimg.cc/fJ1t1Khw)

The linear model for the data is:
$$begin{aligned}
y = 69.789 -0.5349x\
end{aligned}$$

**Part c**
The intercepts of the model are:
$$begin{aligned}
beta_0 = 69.789 text{ and } beta_1=-0.5349\
end{aligned}$$

Meaning,
– The first intercept cannot be interpreted since the value of $x$ will never be equal to zero.
– The stadium revenue decreases by $-0.5349$ for every thousand seat increase in the stadium capacity.

No, the interpretation of the intercepts doesn’t make sense. Usually, establishments are created with a huge seating capacity because they can generate larger revenues than small seating capacity places.

**Part d**
From the data, the university should build a stadium with a capacity of $32$ thousand seats because it produces the largest revenue, $54.7$ million dollars.

**Part a**
Using a graphing calculator, plot the points to create a scatterplot.
|$x$ |$0$ |$15$ |$30$ |$240$ |
|–|–|–|–|–|
|$y$ |$200$ |$140$ |$105$ |$70$ |

Graph:
[](https://postimg.cc/219VD7Wc)

Based on the graph, a linear function can be used.

**Part b**
Scatter plot with line regression line.
[](https://postimg.cc/GTBRwVWz)

The linear model for the data is:
$$begin{aligned}
y = 155.91 -0.3812x\
end{aligned}$$

The intercepts of the model are:
$$begin{aligned}
beta_0 = 155.91 text{ and } beta_1=-0.3812\
end{aligned}$$

Meaning,
– When the tea is poured instantly $(t=0)$, the temperature is equal to $155.91^circ F$.
– The temperature of the tea decreases by $0.3812$ for every minute increase in the time.

The objective of this problem is to determine whether Preethi and Andres will be able to finish $1000$ origami cranes in 6 days given the conditions stated in the problem.

Let’s assume that both of them will work $6$ hours per day for 6 days.

Preethi can make $20$ cranes per hour for **two hours** before resting her hands for 15 minutes and work again. The number of cranes that she can make for one day is.

Preeti Work Done in a day.
$$begin{aligned}
20medspacetext{cranes per hour x $6$ hours}&=120medspacetext{cranes per day}
end{aligned}$$

Since Preethi has 15 minute break for every two hours of work, the total time of her break after working for 6 hours is **30 minutes**.
$$begin{aligned}

end{aligned}$$

For 6 days, the number of cranes she can finish is.
$$begin{aligned}
120medspacetext{cranes x $6$ days}&=720medspacetext{cranes}
end{aligned}$$

For Andres, let’s assume that after his **3 hours of work**, he will rest his hand for **30 minutes** and then work again for 3 hours. Based on the given cranes per hour, the number of cranes he can make for one day is.
$$begin{aligned}
6+8+12&=26medspacetext{cranes(3 hours)}cdot2&=52medspacetext{cranes per day}\

end{aligned}$$

For 6 days, the number of cranes he can finish is.
$$begin{aligned}
52medspacetext{cranes x $6$ days}&=312medspacetext{cranes}
end{aligned}$$

The total cranes that they can finish in 6 days is.
$$begin{aligned}
720+312&=1032medspacetext{cranes}
end{aligned}$$

Therefore, both of them can make $1000$ origami cranes for their teacher before they return to school.

The problem requires modeling the number of cranes made in real-time by persons $P$ and $A$.

**A**
Start by making a table of points for person $P$ given the data in the previous exercise. Label the number of hours with the variable $x$ and the numbers of manufactured cranes with the variable $y$.

$$begin{array}{|c|c|}
hline
x & y \
hline
0 & 0\
hline
1 & 20\
hline
2 & 40\
hline
2.25 & 40\
hline
3.25 & 60\
hline
4.25 & 80\
hline
4.5 & 80\
hline
5.5 & 100\
hline
6.5 & 120\
hline
end{array}$$

Use the linear regression to obtain the linear model for person $P$.

the linear model describing the work of person $P$ is described by the function $f(x)=18.14x+1.03$.

Finally, calculate the number of cranes made by person $P$ after five hours using the obtained model. In other words, calculate $f(5)$.
$$begin{aligned}
f(5)&=18.14cdot5+1.03\
&=91.73\
&approx92
end{aligned}$$

**B**
Start by making a table of points for person $A$ given the data in the previous exercise. Label the number of hours with the variable $x$ and label the numbers of manufactured cranes in that time with the variable $y$. Make an assumption that person $A$ will take a $30$ minutes break after $3$ hours as person $A$ is more inexperienced in making cranes than person $P$. Also, make an assumption that person $A$ is a quick learner and will be able to continue making $12$ cranes per hour after the break.

$$begin{array}{|c|c|}
hline
x & y \
hline
0 & 0\
hline
1 & 6\
hline
2 & 14\
hline
3 & 26\
hline
3.5 & 26\
hline
4.5 & 38\
hline
5.5 & 50\
hline
6.5 & 62\
hline
end{array}$$

Use the graphing calculator to present the points in the coordinate plane.


Observe that the model for person $A$ is also linear.

Use the linear regression to obtain the linear model for person $A$.


The linear model describing the work of person $A$ is $g(x)=9.58x-3.38$.

**C**
My model was made under the assumption that person A is a quick learner and that he will be able to continue manufacturing $12$ cranes per hour after the first three hours. Also, I didn’t want to assume that he will improve much more than that because person P is an experienced crane maker and he does not exceed making $20$ cranes per hour. Therefore, this assumption about the efficiency of person A after three hours seems the most logical to me.

If person $A$ operates under that assumption, then after $5$ hours he will be able to manufacture the number of cranes equal $g(5)$.
$$begin{aligned}
g(5)&=9.58cdot5-3.38\
&=44.52\
&approx45
end{aligned}$$

**D**

Use the obtained models to determine the time needed to make $1000$ cranes. In other words, determine the value of $x$ for which the statement $f(x)+g(x)=1000$.
$$begin{aligned}
f(x)+g(x)&=1000\
18.14x+1.03+9.58x-3.38&=1000\
27.72x&=1000-1.03+3.38\
27.72x&=1002.35\
x&=36.15\
x&approx36
end{aligned}$$
According to the models, it will take them approximately $36$ hours to finish the project of making $1000$ cranes.
The accuracy of this prediction is based upon the assumptions made about the efficiency of the work performed by person $A$.

**E**
In my opinion, this is not a reasonable amount of time to work in $6$ days during the spring break because the spring break should be a time to relax, not manufacturing paper cranes for half of the day.

**Part a**
The number of *petit fours* the chef can make from all the fillings is:
$$begin{aligned}
_8C_3 &= frac{8!}{5! cdot 3!} \\
&= frac{8cdot 7 cdot 6cdot5cdot4cdot3cdot2cdot1}{(5cdot4cdot3cdot2cdot1)(3cdot2cdot1)}\\
&=boxed{56}
end{aligned}$$

**Part b**
The number of *petit fours* with raspberry and custard the chef can make is:
$$begin{aligned}
_6C_1 &= frac{6!}{5! cdot 1!} \\
&= frac{6cdot5cdot4cdot3cdot2cdot1}{(5cdot4cdot3cdot2cdot1)(1)}\\
&=boxed{6}
end{aligned}$$

**Part c**
The number of apricot-filled *petit fours* is:
$$begin{aligned}
_7C_2 &= frac{7!}{5! cdot 2!} \\
&= frac{7cdot6cdot5cdot4cdot3cdot2cdot1}{(5cdot4cdot3cdot2cdot1)(2cdot1)}\\
&= 21
end{aligned}$$

The probability (as a percent) of getting a *petit four* that has apricot
filling is:
$$begin{aligned}
P&= frac{text{number of apricot-filled petit four}}{text{total number of petit four}} times 100 \\
&= frac{_7C_2}{56} times 100\\
&= frac{21}{56} times 100\\
&= 0.375 times 100\\
&= boxed{37.5%}
end{aligned}$$

a. $56$
b. $6$
c. $37.5%$

**Part a**
Since there are no duplicate letters, here’s how to solve it:
$$begin{aligned}
&= 5! \
&= 5 cdot 4 cdot 3 cdot 2 cdot 1 \
&= boxed{120}
end{aligned}$$

**Part b**
Since there is 1 duplicate letter, here’s how to solve it:
$$begin{aligned}
&= frac{5!}{2} \\
&= frac{5 cdot 4 cdot 3 cdot 2 cdot 1}{2} \\
&= frac{120}{2} \\
&= boxed{60}
end{aligned}$$

**Part c**
Since there are 2 duplicate letters, here’s how to solve it:
$$begin{aligned}
&= frac{5!}{2 cdot 2} \\
&= frac{5 cdot 4 cdot 3 cdot 2 cdot 1}{4} \\
&= frac{120}{4} \\
&= boxed{30}
end{aligned}$$

**Concept**
The formula in getting the number of sides of a polygon is:
$$begin{aligned}
s &= (n-2)cdot180 \
end{aligned}$$
where
$s$ = the sum of the interior angles
$n$ = number of sides of a polygon

**Solution**
Using the formula, substitute the values.
$$begin{aligned}
s &= (n-2) cdot 180 \
2160 &= (n-2) cdot 180 \
frac{2160}{180} &= frac{(n-2) cdot 180}{180} \
12+2 &= n-2+2 \
boxed{14} &= n \
end{aligned}$$

**Solution**
Graph $triangle ABC$ first.
[](https://postimg.cc/njxMx4ZN)

Rotate the triangle $90^circ$ clockwise.
[](https://postimg.cc/DJ7fH7PT)

Notice that point $B$ is now on the coordinate $boxed{(7,-3)}$.

**Solution**
Solve the quadratic equations below.
a. $x^2+8 = -6x$
$$begin{aligned}
x^2+8 &= -6x \
x^2+8+6x &= -6x+6x \
x^2+8+6x &= 0 \
(x+2)(x+4) &= 0 \
end{aligned}$$
Thus, $boxed{x=-2}$ and $boxed{x=-4}$

b. $0=3x^2-7x+4$
Get the factors of the equation.
$$begin{aligned}
3x^2-7x+4 &= 0 \
(3x-4)(x-1) &= 0 \
end{aligned}$$
Then solve for $x$ in the first factor,
$$begin{aligned}
3x-4 &= 0 \
3x-4+4 &= 0+4 \
frac{3x}{3} &= frac{4}{3} \
x &= frac{4}{3} \
end{aligned}$$

Thus,
$boxed{x=frac{4}{3}}$ and $boxed{x=1}$

c. $(x+5)(-2x+3)=0$
Solve for $x$ in the second factor.
$$begin{aligned}
-2x+3&=0 \
-2x+3-3&=0-3 \
frac{-2x}{-2} &= frac{-3}{-2} \
x &= frac{3}{2} \
end{aligned}$$

Thus,
$boxed{x= frac{3}{2}}$ and $boxed{x=-5}$

a. $x=-2$ and $x=-4$
b. $x=frac{4}{3}$ and $x=1$
c. $x= frac{3}{2}$ and $x=-5$

**Solution**
Remember that the measure of the intercepted arc is equal to the measure of the central angle.
$$begin{aligned}
text{if } moverset{largefrown}{AB} = 60^circ text{ then, } mangle C = 60^circ
end{aligned}$$

[](https://postimg.cc/cvx3bs5n)

Since $overline{AC}$ is a radius, then it should have the same length as $overline{BC}$.
$$begin{aligned}
overline{AC} = overline{BC}
end{aligned}$$
Thus, $triangle ABC$ is an isosceles triangle with congruent base angles.
$$begin{aligned}
mangle A = mangle B
end{aligned}$$

Then solve for $mangle A$ and $mangle B$.
$$begin{aligned}
mangle A + mangle B + mangle C &= 180^circ \
mangle A + mangle B + 60^circ &= 180^circ \
mangle A + mangle B + 60^circ -60^circ &= 180^circ -60^circ\
mangle A + mangle B &= 120^circ\
end{aligned}$$
Since $mangle A = mangle B$, then divide $120^circ$ to $2$.
$$begin{aligned}
mangle A = mangle B = 60^circ
end{aligned}$$

Notice that all the angles are congruent, then the triangle is equilateral. With that, the length of $overline{AC}$ is $10$.

Now compute for the area of the circle.
$$begin{aligned}
A &= pi r^2 \
&= pi (10)^2 \
&= pi cdot 100 \
&= boxed{314.16 text{ units}^2} \
end{aligned}$$

$A=314.16$ units$^2$

Solve for the volume of the pyramid.
$$begin{aligned}
Volume &= frac{1}{3}bh \
&= frac{1}{3}(81text{ cm}^2)(12text{ cm}) \
&= boxed{324 text{ cm}^3}
end{aligned}$$

Solve for the height.
$$begin{aligned}
V &= frac{1}{3}bh \
12 &= frac{9}{3} cdot h \
12 &= 3h \
frac{12}{3} &= frac{3h}{3} \
boxed{4 text{ cm}} &= h
end{aligned}$$

**Solution**
a. The longest time recorded was the last throw with $t=1.25s$, and the ball has still not reached the ground. Thus, the **last throw** is when the ball stayed in the air longer.

b. Compute for the velocity.
$$begin{aligned}
overline{v} &= frac{triangle text{height}}{triangle text{time}} \
&= frac{5.5-1.5}{0.25-0} \
&= frac{4}{0.25} \
&= boxed{16 ft/s} \
end{aligned}$$

a. last throw
b. $16 ft/s$
c. second throw

**Solution**
Since the batting orders of nine starting baseball players don’t have any conditions, it can be solved using simple factorial.
$$begin{aligned}
&= 9! \
&= 9cdot8cdot7cdot6cdot5cdot4cdot3cdot2cdot1 \
&= boxed{362,880}
end{aligned}$$

$9! = 362,880$

**Solution**
a. Multiply the number of Beethoven symphonies to Mozart concerto to get the number of ways the local radio station KALG can play two pieces.
$$begin{aligned}
9 cdot 27 = boxed{243 text{ ways}}
end{aligned}$$

b. Combine $15$ Schubert string quartet to the ways computed above.
$$begin{aligned}
9 cdot 27 cdot 15 = boxed{3645 text{ ways}}
end{aligned}$$

a. $243$ ways
b. $3645$ ways

**Concept**
Exponential Functions are expressed as:
$$begin{aligned}
y &= i(1 pm r)^t
end{aligned}$$
where:
$y =$ final amount
$i =$ initial amount
$r =$ rate (in percent)
$1 pm r =$ multiplier
$t =$ time

**Solution**
a. Based on the discussion above, the starting point is $20$ and the multiplier is $1.06$.
|$x$ |$-1$ |$0$ |$1$ |
|–|–|–|–|
|$y$ |$19.87$ |$20$ |$21.2$ |

Sketch the graph using the points from the table of values.
[](https://postimg.cc/8J6rxwQF)

b. This function can be described as the growth of the population of an area with only $20$ initial number of people living within a period of time. Specifically, the growth rate of the population is $6%$.

**Solution**
Equate the equations to each other to solve for the values of $x$.
$$begin{aligned}
2x+9 &= x^2-2x-3 \
2x+9 – x^2+2x+3 &= x^2-2x-3+ (-x^2+2x+3) \
– x^2+4x+11 &= 0 \
end{aligned}$$
factoring out,
$$begin{aligned}
– x^2+4x+11 &= 0 \
(x+2)(x-6) &= 0
end{aligned}$$
Thus, $boxed{x=-2}$ and $boxed{x=6}$.

To solve for the values of $y$, substitute the values of $x$.
For $x=-2$,
$$begin{aligned}
y &= 2x+9 \
&= 2(-2)+9 \
&= -4+9 \
&= boxed{5} \
end{aligned}$$

For $x=6$,
$$begin{aligned}
y &= 2x+9 \
&= 2(6)+9 \
&= 12+9 \
&= boxed{21} \
end{aligned}$$

$x=-2$
$x=6$
$y=5$
$y=21$

**Solution**
a. Since there are $5$ inscribed angles in the circle, the measure of angle $a$ is computed by dividing the measurement of the whole circle into $5$. Then, divide by $2$.
$$begin{aligned}
mangle a &= frac{360^circ}{5} \\
&= 72^circ\\
&= frac{72^circ}{2}\\
&= boxed{36^circ}\
end{aligned}$$

b. Measure of angle $b$ is computed by:
$$begin{aligned}
mangle b &= 180^circ – 72^circ \
&= boxed{108^circ}
end{aligned}$$

Measure of angle $c$ can be computed as:
$$begin{aligned}
mangle c &= mangle b \
108^circ &= 108^circ \
&= boxed{108^circ}
end{aligned}$$

Since the supplementary angle of $d$ is the same size as $b$, the measurement of angle $d$ is computed by:
$$begin{aligned}
mangle d &= 180^circ – 108^circ \
&= boxed{72^circ}
end{aligned}$$

**Concept**
The volume of a rectangular prism is computed as:
$$begin{aligned}
V = lcdot wcdot h
end{aligned}$$

b. Compute the volume of the two inner rectangular prisms.
$$begin{aligned}
V_{inner} &= lcdot wcdot h \
&= 0.25 text{ ft} cdot 1 text{ ft} cdot 0.2 text{ ft}\
&= 0.05 text{ ft}^3
end{aligned}$$
Multiply by $2$ since there are two prisms.
$$begin{aligned}
V_{inner} &= 0.05 text{ ft}^3 \
&= 2 times 0.05 text{ ft}^3 \
&= 0.1 text{ ft}^3
end{aligned}$$

Compute for the volume of the outer rectangular prism.
$$begin{aligned}
V_{outer} &= lcdot wcdot h \
&= 1.5 text{ ft} cdot 1 text{ ft} cdot 1 text{ ft}\
&= 1.5 text{ ft}^3
end{aligned}$$

The volume of the block is computed by subtracting the volume of the two inner rectangular prisms from it.
$$begin{aligned}
V &= V_{outer} – V_{inner} \
&= 1.5 text{ ft}^3 – 0.1text{ ft}^3 \
&= boxed{1.4 text{ ft}^3}
end{aligned}$$

**Solution**
a. To solve for the area and perimeter of the resulting triangle, multiply the previous results to $2$.
$$begin{aligned}
A_1 &= A_0 cdot 2 \
&= 36 text{ units}^2 cdot 2 \
&= boxed{72 text{ units}^2}
end{aligned}$$

To solve for the area and perimeter of the resulting triangle, multiply the previous results to $2$.
$$begin{aligned}
P_1 &= P_0 cdot 2 \
&= 42 text{ units} cdot 2 \
&= boxed{84 text{ units}}
end{aligned}$$

b. To get the $frac{1}{3}$ area and perimeter of the resulting triangle, divide the answers from part a to $3$.
$$begin{aligned}
A_2 &= A_1 cdot frac{1}{3} \
&= 72 text{ units}^2 cdot frac{1}{3} \
&= boxed{24 text{ units}^2}
end{aligned}$$

To get the $frac{1}{3}$ area and perimeter of the resulting triangle, divide the answers from part a to $3$.
$$begin{aligned}
P_2 &= P_1 cdot frac{1}{3} \
&= 84 text{ units} cdot frac{1}{3} \
&= boxed{28 text{ units}}
end{aligned}$$

a. $A=72 text{ units}^2$ and $P=84 text{ units}$
b. $A=24 text{ units}^2$ and $P=28 text{ units}$

Based on **figure a**, to get the value of x, the square of 1 unit is created first. Then the line was drawn from midpoint of one side to the corner of the other side. Since the right triangle was formed, the length of line **(hypotenuse)** can be solve by Pythagorean Theorem.
$$begin{aligned}
c^2&=a^2+b^2\
end{aligned}$$

Solving the value of hypotenuse.
$$begin{aligned}
c^2&=1^2+left(dfrac{1}{2}right)^2\
c^2&=dfrac{5}{4}\
c&=sqrt{dfrac{5}{4}}\
c&=sqrt{dfrac{5}{(2)^2}}\
c&=dfrac{sqrt{5}}{2}
end{aligned}$$

Based on figure(b), the value of $x$ can now be solve.
$$begin{aligned}
x&=dfrac{1}{2}+dfrac{sqrt{5}}{2}\
x&=dfrac{1+sqrt{5}}{2}approx1.618
end{aligned}$$

To prove that smaller and bigger rectangle are similar, their ratio with respect to dimensions should be **similar also**. Take note that the ratio for bigger rectangle is $1.618$ which is shown in **Step 5**.

The ratio of the smaller rectangle is.
$$begin{aligned}
text{ratio}&=dfrac{text{larger dimension}}{text{smaller dimension}}\
text{ratio}&=dfrac{1}{dfrac{sqrt{5}}{2}-dfrac{1}{2}}\
text{ratio}&=dfrac{1}{dfrac{sqrt{5}-1}{2}}\
text{ratio}&=1cdot{dfrac{2}{sqrt{5}-1}medspacetext{(Transposition Method)}}\
text{ratio}&=dfrac{2}{sqrt{5}-1}\
text{ratio}&=1.618
end{aligned}$$

To get the scale factor, the dimensions of the larger rectangle divided by the dimensions of the smaller rectangle.The equation is.
$$begin{aligned}
text{scale factor}&=dfrac{1+x}{1+left(dfrac{sqrt{5}}{2}-dfrac{1}{2}right)}\
text{scale factor}&=dfrac{1+left(dfrac{sqrt{5}}{2}+dfrac{1}{2}right)}{1+left(dfrac{sqrt{5}}{2}-dfrac{1}{2}right)}\
end{aligned}$$

Solving the equation.
$$begin{aligned}
text{scale factor}&=dfrac{2.618}{1.618}\
text{scale factor}&=1.618
end{aligned}$$

**b**. The ratio is expressed as exact form.
$$begin{aligned}
text{ratio}&=dfrac{dfrac{sqrt{5}}{2}+dfrac{1}{2}}{1}=dfrac{sqrt{5}}{2}+dfrac{1}{2}\
text{ratio}&=dfrac{1+sqrt{5}}{2}\
end{aligned}$$

The ratio expressed as decimal approximation.
$$begin{aligned}
text{ratio}&=dfrac{1+sqrt{5}}{2}\
text{ratio}&=dfrac{1+2.236}{2}\
text{ratio}&=1.618
end{aligned}$$

Since the required is **golden rectangle** with the golden ratio ($phi$) of $1.618$, the value of $h$ is.
$$begin{aligned}
phi&=dfrac{w}{h}\
1.618&=dfrac{267}{h}\
h&=dfrac{267}{1.618}\
h&approx165medspacetext{cm}
end{aligned}$$

Figure shown in Step 1 was from Exercise 96. To determine whether the rectangle is a golden rectangle, the ratio of it’s dimension should be $1.618$ which is the value of $phi$.

Solving for ratio of rectangle C.
$$begin{aligned}
text{ratio}&=dfrac{1}{dfrac{sqrt{5}}{2}-dfrac{1}{2}}\
text{ratio}&=dfrac{1}{0.618}\
text{ratio}&=1.618=phi
end{aligned}$$

To determine whether the rectangle E is a golden rectangle, it’s ratio must be equal to $phi$ or $1.618$. The dimensions shown in Step 5 is converted to it’s decimal form.

Take note that rectangle D is a square with dimensions of $0.618$ x $0.618$. The midpoint dimension is $0.309$. Solving for the line from midpoint to the opposite corner point using Pythagorean Theorem.
$$begin{aligned}
c^2&=a^2+b^2\
c^2&=0.618^2+0.309^2\
c^2&=0.477\
c&=sqrt{0.477}\
c&=0.691
end{aligned}$$

Solving for the ratio of dimensions of rectangle E.
$$begin{aligned}
text{ratio}&=dfrac{0.618}{0.691-0.309}\
text{ratio}&=dfrac{0.618}{0.382}\
text{ratio}&=1.618=phi
end{aligned}$$

The problem requires answering a series of questions regarding the golden ratio, $phi$.

**A**
Calculate the radius of the right decagon if the length of its side is $1$.
Keep in mind that there are two radii of the right decagon:
1. Incircle radius, $r_1=frac{1}{2}sqrt{5+2sqrt5}cdot a$

2. Circumcircle radius, $r_o=frac{1+sqrt5}{2}cdot a$,
where the parameter $a$ represents the length of the side for the right decagon.

Calculate each if $a=1$.
$$begin{aligned}
r_i&=frac{1}{2}sqrt{5+2sqrt5}cdot 1\
&=1.534\
\
r_o&=frac{1+sqrt5}{2}cdot 1\
&=1.618
end{aligned}$$
Conclude that the circumcircle radius or the radius of the outer circle of this decagon has the radius that is equal to the golden ratio, $phi$.

**B**
Calculate the angles of the golden triangle inside this right decagon.
We know that the $r_o=1.618$ for this right decagon.
Use the formula $r_o=frac{a}{2sin(alpha/2)}$, where $a$ represents the side length of the decagon (and the base of the golden triangle), whereas $alpha$ represents the angle of the golden triangle in the middle of the decagon.
$$begin{aligned}
r_o&=dfrac{a}{2sin(alpha/2)}\
1.618&=dfrac{1}{2sin(alpha/2)}\
2sin(alpha/2)cdot1.618&=1\
3.236sin(alpha/2)&=1\
sin(alpha/2)&=frac{1}{3.236}\
frac{alpha}{2}&=sin^{-1}{frac{1}{3.236}}\
frac{alpha}{2}&=18°\
alpha&=boxed{36°}
end{aligned}$$

Now, we know that any triangle inscribed inside a right polygon is an ISOSCELES TRIANGLE, meaning that the angles adjacent to the base of the triangle are equal. In other words, $alpha+2beta=180°$. Use this information to calculate the value of the angle $beta$.
$$begin{aligned}
alpha+2beta&=180°\
36°+ 2beta&=180°\
2beta&=180°-36°\
2beta&=144°\
beta&=boxed{72°}
end{aligned}$$

**C**
Check the ratio of $overline{AC} text{ to }overline{BC}$ and the ratio of $overline{AB} text{ to }overline{BD}$ in the given pentagram.

Use the ruler to measure the length of the lines in question and then divide them to get their ratio. Then determine if the obtained ratios correspond to the golden ratio.

**D**
Try to calculate the following expression as close as possible.
$sqrt{1+sqrt{1+sqrt{1+sqrt{1+sqrt{1+}sqrt{1+sqrt{1+}cdots}}}}}$

First, calculate $sqrt1$. The result is $1$.

Then, calculate $sqrt{1+sqrt1}$.
$$begin{aligned}
sqrt{1+sqrt1}&=sqrt{1+1}\
&=sqrt2\
&=1.41
end{aligned}$$

Calculate $sqrt{1+sqrt{1+sqrt1}}$.
$$begin{aligned}
sqrt{1+sqrt{1+sqrt1}}&=sqrt{1+sqrt{1+1}}\
&=sqrt{1+sqrt{2}}\
&=sqrt{1+1.414}\
&=sqrt{2.414}\
&=1.55
end{aligned}$$

Now skip a few and proceed to calculate $sqrt{1+sqrt{1+sqrt{1+sqrt{1+sqrt{1+sqrt1}}}}}$
$$begin{aligned}
&sqrt{1+sqrt{1+sqrt{1+sqrt{1+sqrt{1+sqrt1}}}}}\
&=sqrt{1+sqrt{1+sqrt{1+sqrt{1+sqrt{1+1}}}}}\
&=sqrt{1+sqrt{1+sqrt{1+sqrt{1+sqrt{2}}}}}\
&=sqrt{1+sqrt{1+sqrt{1+sqrt{1+1.41}}}}\
&=sqrt{1+sqrt{1+sqrt{1+sqrt{2.41}}}}\
&=sqrt{1+sqrt{1+sqrt{1+1.552}}}\
&=sqrt{1+sqrt{1+sqrt{2.552}}}\
&=sqrt{1+sqrt{1+1.597}}\
&=sqrt{1+sqrt{2.597}}\
&=sqrt{1+1.612}\
&=sqrt{2.612}\
&=1.616
end{aligned}$$

Conclude that the more the expression diverges to infinity, the result converges to the golden ratio.

**E**

Observe the properties of $phi$.

First, determine what happens with the inverse of $phi$.

Let’s examine the ratio of two high Fibonacci that gives a good approximation of the golden ratio, for example, $55$ and $89$.

Their ratio is $frac{89}{55}=1.61818$.

Now, take the inverse of this number.
$$begin{aligned}
dfrac{1}{1.61818}=0.618
end{aligned}$$
Therefore, we get the relation $1+dfrac{1}{phi}=phi$.

This is not true for other irrational numbers. Examine $piapprox3.14$.
$$begin{aligned}
1+frac{1}{pi}approx1.32\
pine1.32
end{aligned}

Now, observe the squared value of $phi$.

$$begin{aligned}
phi^2&=1.61818cdot1.61818\
phi^2&=2.618\
end{aligned}$$
Therefore, we get the relation $1+phi=phi^2$.

This relation is also not true for other irrational numbers.
$$begin{aligned}
1+pi&=pi^2\
1+3.14&=9.87\
4.14&ne9.87
end{aligned}$$

**F**
Solve the system of equations graphically.

Make a table of values for the first equation.
$$begin{array}{|c|c|}
hline
x & y \
hline
-3 & 8\
hline
-2 & 3\
hline
-1 & 0\
hline
0 & -1\
hline
1 & 0\
hline
2 & 3\
hline
3 & 8\
hline
end{array}$$

Make a table of values for the second equation.

$$begin{array}{|c|c|}
hline
x & y \
hline
-3 & -3\
hline
-2 & -2\
hline
-1 & -1\
hline
0 & 0\
hline
1 & 1\
hline
2 & 2\
hline
3 & 3\
hline
end{array}$$

Solve the equations algebraically by substituting the variable $y$ with the variable $x$ in the first equation.
$$begin{aligned}
x&=x^2-1\
x^2-x-1&=0
end{aligned}$$
Solve the quadratic equation using the Quadratic Formula.
$$begin{aligned}
a=1, b=-1, c=-1
end{aligned}$$
$$begin{aligned}
x_{1,2}&= frac{-bpmsqrt{b^2-4ac}}{2a}\
&= frac{-(1)pmsqrt{(-1)^2-4cdot1cdot(-1)}}{2cdot1}\
&= frac{1pmsqrt{1+4}}{2}\
&= frac{1pmsqrt{5}}{2}\
x_1&=frac{1+sqrt{5}}{2} \
x_1&approx 1.618\
x_2&approx-0.618
end{aligned}$$

Then, from the second equation,
$$begin{aligned}
y_1&=1.618\
y_2&=-0.618
end{aligned}$$

Conclude that the solutions to the given system of inequalities are $A(phi, phi)$ and $left(-frac{1}{phi}, -frac{1}{phi} right)$.

**Solution**
License plates of the first type can be computed as:
$$begin{aligned}
T_1&= 26 cdot 26 cdot 26 cdot 10 cdot 10 cdot10 \
&= 17,576,000 \
end{aligned}$$

License plates of the second type can be computed as:
$$begin{aligned}
T_2 &= 9 cdot 26 cdot 26 cdot 26 cdot 10 cdot 10 cdot10 \
&= 158,184,000 \
end{aligned}$$

How many more license plates of the second type are possible?
$$begin{aligned}
T &= 158,184,000-17,576,000 \
&= boxed{140,608,000 text{ more plates}}
end{aligned}$$

License plates containing **INT 2** can be computed as:
$$begin{aligned}
T &= 9 cdot 1 cdot 1 cdot 1 cdot 1 cdot 10 cdot10 \
&= 900 \
end{aligned}$$

The probability of getting license plates containing **INT 2** can be computed as:
$$begin{aligned}
P(INT2) &= frac{900}{158,184,000} \\
&= boxed{frac{1}{175,760}}\
end{aligned}$$

**Solution**
a. The number of *petit four* they can make is computed as:
$$begin{aligned}
&= _{12}C_4 + _{12}C_3 \\
&= frac{12!}{8! cdot 4!} + frac{12!}{9! cdot 3!} \\
&= 495 + 220\
&= boxed{715}
end{aligned}$$

b. The number of *petit four* that has both raspberry and custard filling is:
$$begin{aligned}
&= _{10}C_2 \
&= frac{10!}{8! cdot 2!} \
&= 45
end{aligned}$$

The probability of getting a *petit four* that has both raspberry and custard filling is:
$$begin{aligned}
P &= frac{_{10}C_2}{715} times 100\\
&= frac{45}{715} times 100 \\
&= 0.0629 times 100 \
&= boxed{6.29%}
end{aligned}$$

Solve for the volume of the bigger pyramid.
$$begin{aligned}
V_1 &= frac{1}{3}bh \
&= frac{1}{3}(81text{ cm}^2)(12text{ cm}) \
&= boxed{324 text{ cm}^3}
end{aligned}$$

Solve for the volume of the smaller pyramid.
$$begin{aligned}
V_2 &= frac{1}{3}bh \
&= frac{1}{3}(9text{ cm}^2)(3text{ cm}) \
&= boxed{9 text{ cm}^3}
end{aligned}$$

Get the difference of their volumes.
$$begin{aligned}
V &= V_1-V_2 \
&= 324text{ cm}^3- 9 text{ cm}^3 \
&= boxed{315 text{ cm}^3}
end{aligned}$$

$315 text{ cm}^3$

**Concept**
The equation of a circle can be described by the following:
$$begin{gather}
(x-h)^2 + (y-k)^2 = r^2
end{gather}$$

**Given**
We are given with the following system of equations:
$$begin{cases}
x^2+ y^2 = 20 \
(x-6)^2 + y^2 = 32
end{cases}$$

We obtain the value of $y^2$ from the first equation as follows:
$$begin{aligned}
x^2 + y^2 &= 20 \
y^2 &= 20 -x^2
end{aligned}$$

We use the value of $y^2$ obtained from the first equation and plug it to the second equation and then solve for $x$ as follows:
$$begin{aligned}
(x-6)^2 + y^2 &= 32 \
y^2 &= 20 -x^2 \
(x-6)^2 + (20-x^2) &= 32 \
(x^2 -12x + 36) + (20-x^2) &= 32 \
-12x + 56 &= 32 \
-12x &= 32 – 56 \
x &= frac{32-56}{-12} \
&= boxed{2}
end{aligned}$$

Next, we determine the value of $y$ using the value of $x$ as follows:
$$begin{aligned}
x^2 + y^2 &= 20 \
x &= 2 \
y^2 &= 20 – x^2 \
y^2 &= 20 – (2)^2 \
y^2 &= 16 \
y &= boxed{pm 4}
end{aligned}$$

We observe that from the $x$ and $y$ values, the two equations will intersect at the points $(2,-4)$ and $(2,4)$. Hence, there will be two points of intersection.

**Given**
We know that there are $52$ playing cards in a deck. Dillon was able to select $4$ cards from the deck (including one ace). Determine the probability of getting an ace for the fifth card.

We obtain the probability of the scenario by getting the number of aces left over the number of cards left. We calculate for these values as follows:
$$begin{aligned}
A &= 4 – 1 \
A &= boxed{3} \
B &= 52 – 4 \
B &= boxed{48}
end{aligned}$$

We divide the value of $B$ to $A$ to determine the probability of getting an ace as follows:
$$begin{aligned}
&= frac{A}{B} \
&= frac{3}{48} \
&= frac{1}{16}
end{aligned}$$

There probability is $boxed{text{A.} frac{1}{16}}$

$text{A.} frac{1}{16}$

Based on the given equation in the problem, the equation is a **quadratic equation**. The standard form of quadratic equation is.
$$begin{aligned}
ax^2+bx+c=0
end{aligned}$$

where

$a, b, c$ – are the coefficients.

The coefficients based on the given equation in the problem are.
$$begin{aligned}
a&=5\
c&=20\
end{aligned}$$

To determine whether the value of $b$ does have real solution/s, the **discriminant in the quadratic formula** will be used.
$$begin{aligned}
b^2-4ac&=0&text{textcolor{#4257b2}{discriminant}}
end{aligned}$$

Substituting the values of $a$ and $c$ by using the **discriminant** in the quadratic formula.
$$begin{aligned}
b^2-4ac&=0\
b^2&=4ac&text{textcolor{#4257b2}{Transposition Method}}\
b^2&=4(5)(20)\
b^2&=400\
b&=sqrt{400}\
b&=20
end{aligned}$$

To have real solutions in a quadratic equation, the value of $b$ must be,
$$begin{aligned}
b&>20medspacetext{(two real solutions)}\
b&=20medspacetext{(one real solution)}\
b&<20medspacetext{(no real solution)}\
end{aligned}$$

Therefore, to have real solution/s, the value of $b$ must be.
$$begin{aligned}
bgeq20
end{aligned}$$

The circumference and area of a circle has the following formula shown below.

Let

$x$ = circumference of the circle
$A$ = area of the circle
$r$ = radius of the circle
$d$ = diameter of the circle
$$begin{aligned}
x&=2pi r\
A&=pi r^2
end{aligned}$$

To solve for the circumference and area, the value of $r$ must be obtained. Take a look at the figure shown in Step 1. The line $overline{AF}$ is the **diameter of the circle** and is also equal to the line $overline{DE}$. Remember that the **diameter** of the circle is **twice the radius**. Therefore, the equation can be written as,
$$begin{aligned}
overline{AF}=overline{DE}=2r
end{aligned}$$

As shown in the diagram, the triangle $triangle{ACX}$ is formed. The value of one side is,
$$begin{aligned}
r&=overline{CX}+2\
r-2&=overline{CX}
end{aligned}$$

By using the Law of Cosines to get the value of $theta$.
$$begin{aligned}
(overline{CX})^2&=r^2+4^2-2(r)(4)costheta\
(r-2)^2&=r^2+16-8rcostheta\
r^2-4r+4&=r^2+16-8rcostheta\
r^2-4r+4-r^2-16&=-8rcosthetaenspacetext{textcolor{#4257b2}{(Transposition Method)}}\
-4r-12&=-8rcostheta\
-4(r+3)&=-8rcostheta
end{aligned}$$

Divide both sides of the equation by $-4$.
$$begin{aligned}
dfrac{-4(r+3)}{-4}&=dfrac{-8rcostheta}{-4}\
r+3&=2rcostheta\
dfrac{r+3}{2r}&=costheta\
cos^{-1}left(dfrac{r+3}{2r}right)&=theta
end{aligned}$$

When we observe any circle, the **peripheral angle** that corresponds to the diameter is **right angle**. Therefore, right triangle $triangle{AFB}$ is formed as shown in the diagram in **Step 1**.

Solving the value of $theta_1$ in right triangle $triangle{AFB}$.
$$begin{aligned}
theta_1&=180degree-90degree-theta\
theta_1&=90degree-cos^{-1}left(dfrac{r+3}{2r}right)
end{aligned}$$

Applying the **trigonometric ratios** on $theta_1$. Take note that line $overline{AB}=7$
$$begin{aligned}
sintheta_1&=dfrac{overline{AB}}{overline{AF}}\
sinleft[90degree-cos^{-1}left(dfrac{r+3}{2r}right)right]&=dfrac{7}{2r}
end{aligned}$$

Using **trigonometric conversions** on left side of the equation.
$$begin{aligned}
sin(A-B)&=sin Acos B-cos Asin B
end{aligned}$$

$$begin{aligned}
sinleft[90degree-cos^{-1}left(dfrac{r+3}{2r}right)right]
&=sin90degreecosleft[cos^{-1}left(dfrac{r+3}{2r}right)right]-cos90degreesinleft[cos^{-1}left(dfrac{r+3}{2r}right)right]\
&=1cdot left(dfrac{r+3}{2r}right)-0cdotsinleft[cos^{-1}left(dfrac{r+3}{2r}right)right]\
&=dfrac{r+3}{2r}
end{aligned}$$

Substituting the value obtained in **Step 10** into the equation in **Step 9**.
$$begin{aligned}
sinleft[90degree-cos^{-1}left(dfrac{r+3}{2r}right)right]&=dfrac{7}{2r}\
dfrac{r+3}{2r}&=dfrac{7}{2r}
end{aligned}$$

Multiply both sides of the equation by $2r$.
$$begin{aligned}
dfrac{r+3}{2r}cdot(2r)&=dfrac{7}{2r}cdot(2r)\
r+3&=7\
end{aligned}$$

Subtract both sides of the equation by $3$.
$$begin{aligned}
r+3-textcolor{#4257b2}{3}&=7-textcolor{#4257b2}{3}\
r&=4
end{aligned}$$

Solving for the **circumference** of the circle.
$$begin{aligned}
x&=2pi r\
x&=2(3.1416)(4)\
x&=25.133
end{aligned}$$

Solving for the **area** of the circle.
$$begin{aligned}
A&=pi r^2\
A&=(3.1416)(4)^2\
A&=50.266
end{aligned}$$

**A**
Determine the value of the parameters $a, b$ if it is given that the exponential function passes through the given points. Substitute the variables $x$ and $f(x)$ from the given equation with the coordinates of the given points to get the system of equations.
$$
left{
begin{array}{cc}
6=ab^1 \
30.375=ab^5
end{array}
right.$$
Divide the second equation by the first equation to obtain the value of the parameter $b$.
$$begin{aligned}
frac{30.375}{6}&=frac{ab^5}{ab}\
b^4&=frac{81}{16}\
b&=pmfrac{3}{2}
end{aligned}$$
Calculate the possible value of the parameter $a$ using the first equation.
$$begin{aligned}
acdotfrac{3}{2}&=6\
a&=6cdotfrac{2}{3}\
a_1&=4\
\
acdotfrac{-3}{2}&=6\
a&=6cdotfrac{-2}{3}\
a_2&=-4\
end{aligned}$$

Therefore, the possible exponential functions that pass through the given points are:
– $f(x)=4cdotleft(dfrac{3}{2}right)^x$

– $f(x)=-4cdotleft(-dfrac{3}{2}right)^x$

Beware! The base of an exponential function $(b)$ has to be a **POSITIVE** real number. Therefore, the expression under the number two is **NOT** and exponential function because its base is not a positive number.

**B**
The parameter $b$ is the base of the exponential function. It tells is us on which side of the coordinate plane the graph of the exponential function is located, relative to the $x$-axis. If the base of the exponential function is a positive real number greater than $1$, the exponential function is growing, whereas the exponential function is decaying if the value of the base is in the interval $0<b<1$.

**Given**
A train has $8$ cars comprising of $2$ first-class, $5$ second-class, and $1$ restaurant car. We determine the number of ways these cars can be linked together.

**Solution**
Since there are a total of $8$ cars, we can obtain the number of ways these cars can be linked as follows:
$$begin{aligned}
8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 &= 8!
end{aligned}$$

We consider that the class cars should be linked together in such a way that the two-first class cars should be beside each other (same way with the second-class car). Hence, we determine the number of ways the first-class cars can be linked as follows:
$$begin{aligned}
2 times 1 &= 2!
end{aligned}$$

Next, we determine the number of ways the second-class cars can be linked as follows:
$$begin{aligned}
5 times 4 times 3 times 2 times 1 &= 5!
end{aligned}$$

We obtain the number of ways these cars can be linked together by dividing the number of ways the cars can be linked (without restriction) by the number of ways the class-specific cars should be linked as follows:
$$begin{aligned}
&= frac{8!}{5! 2!} \
&= frac{40320}{240} \
&= boxed{168}
end{aligned}$$

Hence, there are $168$ ways that we can link the cars.

$168 text{ ways}$

**Part a**
Arik charges $$5.00$ when driving for less than $3$ miles, and charges additional $$1.50$ per mile for driving more than $3$ miles. We determine the piecewise function of Arik’s earning using these information as follows:
$$f(x)=
begin{cases}
5 & 0 < x <3 \
5 + 1.5x & 3 le x
end{cases}$$

where $x$ is the number of miles driven

**Part b**
The price per galloon of gas is $$3.00$ and Arik’s mileage is $28 frac{text{mi}}{text{gal}}$. While his car’s maintenance cost is $$ 0.60$ per mile. We obtain the following function for Arik’s fuel and maintenance cost per mile driven:
$$begin{gather}
f(x)=
-frac{3x}{28} – 0.60x
end{gather}$$
where $x$ is the number of miles driven

**Part c**
We create the new piecewise function that represents Arik’s profit by combining the profit (part a) and cost (part b) functions as follows:
$$f(x)=
begin{cases}
5 & 0 < x 0
end{cases}$$

We determine Arik’s predicted profit when he drives for $2.5$ miles. We consider the profit obtained as the first equation since he drives for less than $3$ miles. We calculate for the profit using the new piecewise function as follows:
$$begin{aligned}
x &= 2.5 \
f(x) &= 5 -frac{3x}{28} – 0.60x \
f(2.5) &= 5 – frac{3cdot 2.5}{28} – 0.60(2.5) \
&= 5 -1.23 \
&= boxed{3.77}
end{aligned}$$

Arik’s predicted profit when he drives for $2.5$ miles is $$3.77$

Next, we determine the profit when the errand requires him to drive for $10$ miles using the piecewise function as follows:
$$begin{aligned}
x &= 10 \
f(x) &= (5+1.5x) -frac{3x}{28} – 0.60x \
f(10) &= (5+1.5 (10)) + left(- frac{3cdot 10}{28} – 0.60(10) right) \
&= (20) + (-7.07) \
&= boxed{13.93}
end{aligned}$$

Arik’s predicted profit when he drives for $10$ miles is $$13.93$

**Part a**
We determine the area of the diagram by separating the diagram into $3$ rectangles and $1$ triangle. We illustrate these sections as follows:
[](https://postimg.cc/G9pkKYX0)

We know that the three rectangles have equal lengths. Upon observing the diagram, the value of the length of the rectangle is equal to the length of the side of the square figure on the top of the diagram. We first take a look at the triangle on the right. We can obtain the value of the height of the rectangle using the trigonometric identity $sin theta = frac{text{opposite}}{text{hypotenuse}}$.

Given that $theta = 60^circ$. We obtain the value of the opposite side as follows:
$$begin{aligned}
sin theta &= frac{text{opposite}}{text{hypotenuse}} \
sin 60^circ &= frac{text{opposite}}{30} \
text{opposite} &= 30 sin (60^circ) \
&= 25.98 \
&approx boxed{26 text{ mm}}
end{aligned}$$

Next, now that we obtained the opposite side, we can calculate for the adjacent side using $cos theta = frac{text{adjacent}}{text{hypotenuse}}$. We calculate for the adjacent side of the triangle as follows:
$$begin{aligned}
cos theta &= frac{text{adjacent}}{text{hypotenuse}} \
cos 60^circ &= frac{text{adjacent}}{30} \
text{adjacent} &= 30 cos (60^circ) \
&= boxed{15 text{ mm}} \
end{aligned}$$

Now that we have obtained the value of the height and base of the triangle, we can obtain its area as follows:
$$begin{aligned}
A_t &= frac{1}{2}bh \
&= frac{1}{2} (15) cdot (26) \
&= boxed{195 text{ mm}^2}
end{aligned}$$

Next, we obtain the length of each rectangle by subtracting the base of the triangle to the overall length of the bottom of the diagram. We calculate for the length as follows:
$$begin{aligned}
L_r &= frac{51 – L_t}{3} \
&= frac{51-15}{3} \
&= boxed{12}
end{aligned}$$

We calculate for the area of the rectangle using the length and height of the triangle as follows:
$$begin{aligned}
A_r &= l cdot h \
&= (12) cdot (26) \
&= boxed{192 text{ mm}^2}
end{aligned}$$

Next, we need to determine the area of the square figure. We know that the length of each side of the square is equal to the length of the rectangle. We calculate for this area as follows:
$$begin{aligned}
A_s &= l^2 \
&= (12)^2 \
&= boxed{144 text{ mm}^2}
end{aligned}$$

To obtain the area of the diagram, we add the area of the $3$ rectangles (assuming that they are all complete), area of the triangle and subtract the area of the square. We calculate for the total area as follows:
$$begin{aligned}
A_{total} &= 3 A_r + A_t – A_s \
&= 3 (192) + (195) – (144) \
&= boxed{627 text{ mm}^2}
end{aligned}$$

The area of the diagram is $627 text{ mm}^2$

**Part b**
If the diagram is enlarged by $400%$, this means that the total area of the diagram is increased by $4$ times. We calculate for the new area as follows:
$$begin{aligned}
A_{new} &= A_{total} cdot 4 \
&= 627 cdot 4 \
&= boxed{2508 text{ mm}^2}
end{aligned}$$

The new area when the diagram is enlarged will be $2508 text{ mm}^2$

a. $627 text{ mm}^2$
b. $2508 text{ mm}^2$

**Solution**
Compute the volume of the cylinder.
[](https://postimg.cc/phzmjw6q)

$$begin{aligned}
V_1 &= pi r^2h \
&= pi cdot (38 text{ units})^2 cdot 71text{ units} \
&= pi cdot 1444 text{ units}^2 cdot 71text{ units} \
&= 322,088.65 text{ units}^3 \
end{aligned}$$

Compute for the volume of the rectangular prism.
[](https://postimg.cc/VrqqjGGB)

$$begin{aligned}
V_1 &= lwh \
&= 84text{ units} cdot 34 text{ units} cdot 99text{ units} \
&= 282,744 text{ units}^3 \
end{aligned}$$

Based on the diagram, the lines $overline{AC}$ and $overline{BC}$ are **congruent** because these lines are the radius of the circle. The triangle $triangle{ACB}$ inscribed in the circle is an isosceles triangle because it has $2$ equal sides. Therefore, the angles opposite the equal sides are also equal which is denoted by angle $x$.

The sum of total angles in a triangle is $180degree$. Therefore, the equation for the angle $x$ will be.
$$begin{aligned}
x+x+theta&=180degree
end{aligned}$$

Take note that central angle $theta$ is equal to the angle of arc $AB$. Therefore, the equation will be.
$$begin{aligned}
x+x+106degree&=180degree\
2x+106degree&=180degree
end{aligned}$$

Subtract both sides of the equation by $106degree$.
$$begin{aligned}
2x+106degree-textcolor{#4257b2}{106degree}&=180degree-textcolor{#4257b2}{106degree}\
2x&=74degree
end{aligned}$$

Divide both sides of the equation by $2$.
$$begin{aligned}
dfrac{2x}{textcolor{#4257b2}{2}}&=dfrac{74degree}{textcolor{#4257b2}{2}}\
x&=37degree
end{aligned}$$

**b**. Based on the diagram, the quadrilateral is a trapezoid. **The angles on the same side between the bases are supplementary.** Therefore, the equation will be.
$$begin{aligned}
x+67degree&=180degree&text{Eq. 1}\
5y+(3y-16degree)&=180degree&text{Eq. 2}
end{aligned}$$

Solving the value of $x$ in Eq. 1 by subtracting $67degree$ on both sides of the equation.
$$begin{aligned}
x+67degree-textcolor{#4257b2}{67degree}&=180degree-textcolor{#4257b2}{67degree}\
x&=113degree
end{aligned}$$

Solving the value of $y$ in Eq. 2.
$$begin{aligned}
5y+3y-16degree&=180degree\
8y-16degree&=180degree
end{aligned}$$

Add $16degree$ on both sides of the equation.
$$begin{aligned}
8y-16degree+textcolor{#4257b2}{16degree}&=180degree+textcolor{#4257b2}{16degree}\
8y&=196degree
end{aligned}$$

Divide both sides of the equation by $8$.
$$begin{aligned}
dfrac{8y}{textcolor{#4257b2}{8}}&=dfrac{196degree}{textcolor{#4257b2}{8}}\
y&=24.5degree
end{aligned}$$

**a**. Based on the diagram, the value of $x$ can be obtain using **Pythagorean Theorem by trigonometric ratio.**

Take note that $sintheta=dfrac{text{opposite}}{text{hypotenuse}}$,
$$begin{aligned}
sin60degree&=dfrac{x}{28}\
0.866&=dfrac{x}{28}\
end{aligned}$$

Multiply both sides of the equation by $28$.
$$begin{aligned}
0.866cdot(28)&=dfrac{x}{28}cdot(28)\
24.249&=x
end{aligned}$$

**b**. Based on the diagram, the value of $x$ **cannot be solve** because **the value of the hypotenuse is less than the value of the opposite side**. The value of hypotenuse in a right triangle **should always be higher than any sides of a right triangle**. Therefore, the triangle does not exist.

**c**. Based on the given values in the triangle, the value of $x$ can be solve by **using the formula of the area of the triangle**.

Take note that area of triangle is.
$$begin{aligned}
A&=dfrac{1}{2}bh
end{aligned}$$

where,
$b$ = base length.
$h$ = height of the triangle.

Solving for the value of $x$ using the formula for area of triangle. Take note that the height of the **shaded triangle** will be $x$. The area for shaded triangle will be denoted by $A_s$.
$$begin{aligned}
A_s&=dfrac{1}{2}cdot (b)cdot(x)\
96&=dfrac{1}{2}cdot(8)cdot(x)\
96&=4x
end{aligned}$$

Divide both sides of the equation by $4$.
$$begin{aligned}
dfrac{96}{textcolor{#4257b2}{4}}&=dfrac{4x}{textcolor{#4257b2}{4}}\
24medspacetext{units}&=x
end{aligned}$$

Solving the length of $overline{AC}$ by using the Pythagorean Theorem trigonometric ratio.
$$begin{aligned}
sinangle{A}&=dfrac{overline{BC}}{overline{AC}}\
sin42degree&=dfrac{7}{overline{AC}}
end{aligned}$$

Multiply both sides of the equation by $overline{AC}$.
$$begin{aligned}
sin42degreecdot(overline{AC})&=dfrac{7}{overline{AC}}cdot(overline{AC})\
sin42degreeoverline{AC}&=7
end{aligned}$$

Divide both sides of the equation by $sin42degree$.
$$begin{aligned}
dfrac{sin42degreeoverline{AC}}{sin42degree}&=dfrac{7}{sin42degree}\
overline{AC}&=10.46medspacetext{mm}
end{aligned}$$

B. $10.5medspacetext{mm}$

The problem requires determining the quadratic equations that pass through the given points.

The standard form of the quadratic equation is $ax^2+bx+c=0$.

Use the $x$-intercept form of the quadratic equations to solve the problem.
$$begin{aligned}x^2-x(x_1+x_2)+ (x_1cdot x_2)=0 end{aligned}$$

**A**
If the $x$-intercepts of this quadratic equation are $x_1=-3$ and $x_2=2$, then:
$$begin{aligned}
x^2-x(x_1+x_2)+ (x_1cdot x_2)&=0\
x^2-x(-3+2)+ (-3cdot2)&=0\
x^2-xcdot(-1)-6&=0\
x^2+x-6&=0
end{aligned}$$

Check your answer by solving the obtained quadratic equation using the Quadratic Formula:
$$begin{aligned}
a=1, b=1, c=-6
end{aligned}$$
$$begin{aligned}
x_{1,2}&= frac{-bpmsqrt{b^2-4ac}}{2a}\
&= frac{-1pmsqrt{1^2-4cdot1cdot(-6)}}{2cdot1}\
&= frac{-1pmsqrt{1+24}}{2}\
&= frac{-1pmsqrt{25}}{2}\
x_{1,2}&= frac{-1pm5}{2}\
x_1&=2 \
x_2&=-3
end{aligned}$$

**B**
If the $x$-intercepts of this quadratic equation are $x_1=-3$ and $x_2=frac{1}{2}$, then:
$$begin{aligned}
x^2-x(x_1+x_2)+ (x_1cdot x_2)&=0\
x^2-x(-3+0.5)+ (-3cdot0.5)&=0\
x^2-xcdotleft(-frac{5}{2}right)+left(-frac{3}{2}right)&=0\
x^2+frac{5}{2}x-frac{3}{2}&=0\
end{aligned}$$

Check your answer by solving the obtained quadratic equation using the Quadratic Formula:
$$begin{aligned}
a=1, b=2.5, c=-1.5
end{aligned}$$
$$begin{aligned}
x_{1,2}&= frac{-bpmsqrt{b^2-4ac}}{2a}\
&= frac{-2.5pmsqrt{2.5^2-4cdot1cdot(-1.5)}}{2cdot1}\
&= frac{-2.5pmsqrt{6.25+6}}{2}\
&= frac{-2.5pmsqrt{12.25}}{2}\
x_{1,2}&= frac{-2.5pm3.5}{2}\
x_1&=frac{1}{2} \
x_2&=-3
end{aligned}$$

**Solution**
a. The number of possible Judicial Boards are:
$$begin{aligned}
_{900}C_{12} &= frac{900!}{888! cdot 12!}\\
&= boxed{5.47758 times 10^{26}}
end{aligned}$$

b. The number of possible Judicial Boards that includes Mariko are:
$$begin{aligned}
_{899}C_{11} &= frac{899!}{888! cdot 11!}\\
&= boxed{7.30343 times 10^{24}}
end{aligned}$$

c. The probability that Mariko will be included in the Judicial Boards is:
$$begin{aligned}
P&= frac{_{899}C_{11} }{_{900}C_{12}} times 100\\
&= frac{7.30343 times 10^{24}}{5.47758 times 10^{26}} times 100\\
&= boxed{1.33%}
end{aligned}$$

a. $5.47758 times 10^{26}$
b. $7.30343 times 10^{24}$
c. $1.33%$

Solve for the volume of the square-based pyramid.
$$begin{aligned}
V &= frac{1}{3} cdot l cdot w cdot h \
&= frac{1}{3} cdot 9 text{ units} cdot 9 text{ units} cdot 48 text{ units} \
&= frac{1}{3} cdot 3888 text{ units}^3\
&= boxed{1296 text{ units}^3}
end{aligned}$$

B. $1296 text{ units}^3$

**Concept**
Remember how to convert a radical expression into its exponential form.
$$begin{aligned}
sqrt[n]{a^x} &= a^{frac{x}{n}}
end{aligned}$$

**Solution**
Convert the following radical expression in exponent form.
a.
$$begin{aligned}
sqrt[3]{10} &= 10^{frac{1}{3}}
end{aligned}$$

b.
$$begin{aligned}
sqrt{15} &= 15^{frac{1}{2}}
end{aligned}$$

c.
$$begin{aligned}
sqrt[4]{18^3} &= 18^{frac{3}{4}}
end{aligned}$$

d.
$$begin{aligned}
frac{1}{sqrt{5}} &= frac{1}{5^{frac{1}{2}}}
end{aligned}$$

Based on the diagram, the triangle $triangle{ADB}$ has **two congruent sides** which is $r$. Therefore, the triangle is an isosceles triangle. An isosceles triangle has **two equal sides** $r$ and **two equal angles** $theta$. The sum of all angles in a triangle is $180degree$.

Take note that the rule for central angle $angle{ADB}$ is.
$$begin{aligned}
angle{ADB}&=overset{largefrown}{AB}\
angle{ADB}&=32degree
end{aligned}$$

Solving for the value of $theta$.
$$begin{aligned}
theta+theta+32degree&=180degree\
2theta+32degree&=180degree
end{aligned}$$

Subtract both sides of the equation by $32degree$.
$$begin{aligned}
2theta+32degree-(32degree)&=180degree-32degree\
2theta&=148degree
end{aligned}$$

Divide both sides of the equation by $2$.
$$begin{aligned}
dfrac{2theta}{2}&=dfrac{148degree}{2}\
theta&=74degree
end{aligned}$$

Applying **Law of Sines** to get the value of $r$.
$$begin{aligned}
dfrac{r}{sintheta}&=dfrac{9}{sinangle{ADB}}\
dfrac{r}{sin74degree}&=dfrac{9}{sin32degree}\
dfrac{r}{sin74degree}&=16.984\
end{aligned}$$

Multiply both sides of the equation by $sin74degree$.
$$begin{aligned}
dfrac{r}{sin74degree}cdot(sin74degree)&=16.984cdot(sin74degree)\
r&=16.326medspacetext{mm}
end{aligned}$$

Using the formula for **Arc Length**.
$$begin{aligned}
length_{overset{largefrown}{AB}}&=dfrac{overset{largefrown}{AB}}{360degree}cdot2pi r
end{aligned}$$

Solving for the arc length $AB$.
$$begin{aligned}
length_{overset{largefrown}{AB}}&=dfrac{32degree}{360degree}cdot2(3.1416)(16.326)\
length_{overset{largefrown}{AB}}&=0.089cdot102.579\
length_{overset{largefrown}{AB}}&=9.129medspacetext{mm}
end{aligned}$$

Note that the parabola is written in its vertex form. The vertex form of the parabola is $y=a(x-h)+k$, where the parameters $h,k$ represent the coordinates of the vertex of the parabola, $V(h,k)$. Therefore, the vertex of this parabola is the point $V(1,-3)$ and that the parabola is facing downwards since there is a minus sign in front of the parentheses.

Make a table of values for the parabola and for the line by selecting a number of values of the variable $x$ and then calculating the value of the variable $y$.
For the parabola:
$$begin{aligned}
f(-2)&=-2(-2-1)^2-3\
&=-2cdot(-3)^2-3\
&=-2cdot9-3\
&=-18-3\
&=-21
end{aligned}$$

The table of values for the parabola is:
$$begin{array}{|c|c|}
hline
x & y \
hline
-2 & -21\
hline
-1 & -11\
hline
0 & -5\
hline
1 & -3\
hline
2 & -5\
hline
3 & -11\
hline
4 & -21\
hline
end{array}$$

For the line:
$$begin{aligned}
f(0)&=-frac{5}{2}cdot0-5\
&=-5
end{aligned}$$

The table of values for the line is:
$$begin{array}{|c|c|}
hline
x & y \
hline
-3 & 2.5\
hline
-2 & 0\
hline
-1 & -2.5\
hline
0 & -5\
hline
1 & -7.5\
hline
2 & -10\
hline
3 & -12.5\
hline
end{array}$$

Connect the dots to get the graphs of both equations.

Conclude that the line intersects the parabola twice, meaning there are $2$ solutions to this system of equations.

$2$

**Solution**
In approximating the value of $x$, $sin(theta)$ will be used since the size computing is the opposite side, and the given one is the hypotenuse.
$$begin{aligned}
sin(theta) &= frac{opp}{hyp}\\
sin(60^circ) &= frac{x}{10}\\
10 cdot 0.469 &= x \
boxed{4.69} &= x
end{aligned}$$

$x approx 4.69$

Pythagorean Theorem is stated as:
$$begin{aligned}
z^2 &= x^2 +y^2\
sqrt{z^2} &= sqrt{x^2 +y^2}\
& boxed{z = sqrt{x^2 +y^2}}
end{aligned}$$

**Solution**
Compute for the number of pizzas the pizza parlor can create with five or fewer toppings.
For the five toppings:
$$begin{aligned}
_{12}C_5 &= frac{12!}{7! cdot 5!} \\
&= 792
end{aligned}$$

For the four toppings:
$$begin{aligned}
_{12}C_4 &= frac{12!}{8! cdot 4!} \\
&= 495
end{aligned}$$

For the three toppings:
$$begin{aligned}
_{12}C_3 &= frac{12!}{9! cdot 3!} \\
&= 220
end{aligned}$$

For the two toppings:
$$begin{aligned}
_{12}C_2 &= frac{12!}{10! cdot 2!} \\
&= 66
end{aligned}$$

For the one topping:
$$begin{aligned}
_{12}C_1 &= frac{12!}{11! cdot 1!} \\
&= 12
end{aligned}$$

For the no topping:
$$begin{aligned}
_{12}C_0 &= frac{12!}{12! cdot 0!} \\
&= 1
end{aligned}$$

Compute for the total number of different ways they can create pizza.
$$begin{aligned}
&= _{12}C_5 + _{12}C_4 +_{12}C_3+_{12}C_2+_{12}C_1+_{12}C_0 \
&= 792 + 495+220 +66+12+1 \
&= boxed{1586}\
end{aligned}$$

There are $1586$ different pizzas they can create with five or fewer toppings.

**Solution**
a. By plotting the points, we can see that the quadrilateral $GHIJ$ is a rhombus. It is a quadrilateral with equal opposite angles, and sides that are are congruent.
[](https://postimg.cc/Lqw8nDpP)

For $overline{GI}$:
$$begin{aligned}
frac{y-y_1}{x-x_1} &= frac{y_2-y_1}{x_2-x_1} \\
frac{y-2}{x+2} &= frac{6-2}{6+2} \\
frac{y-2}{x+2} &= frac{4}{8} \\
x+2 cdot frac{y-2}{x+2} &= frac{4}{8} cdot x+2\\
y-2 &= frac{4(x+2)}{8}\\
y-2 &= frac{4x+8}{8} \\
y &= frac{1}{2}x + 2\\
end{aligned}$$

For $overline{HJ}$:
$$begin{aligned}
frac{y-y_1}{x-x_1} &= frac{y_2-y_1}{x_2-x_1} \\
frac{y-2}{x-3} &= frac{6-2}{1-3} \\
frac{y-2}{x-3} &= frac{4}{-2} \\
x-3 cdot frac{y-2}{x-3} &= -2 cdot x-3\\
y-2 &= -2x+6\\
y &= -2x+8\\
end{aligned}$$

c. The slope of the longer diagonal is greater than the shorter one. The diagonals are perpendicular since they bisect each other, this makes it an orthodiagonal quadrilateral. The sum of their two adjacent angles is $180^circ$, but notice that the angles are all right-angle.

d. If the rhombus is rotated clockwise about $90^circ$, the coordinates of point $J’$ is $(2,-1)$.
[](https://postimg.cc/zLg7VT2Q)

e. Calculate the length of diagonal $overline{GI}$.
$$begin{aligned}
d_1 &= sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \
&= sqrt{(6+2)^2+(6-2)^2} \
&= sqrt{(8)^2+(4)^2} \
&= sqrt{64+16} \
&= sqrt{80} \
&= 8.94
end{aligned}$$

Calculate the length of diagonal $overline{HJ}$.
$$begin{aligned}
d_2 &= sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \
&= sqrt{(1-3)^2+(6-2)^2} \
&= sqrt{(-2)^2+(4)^2} \
&= sqrt{4+16} \
&= sqrt{20} \
&= 4.47
end{aligned}$$

Calculate the area of quadrilateral $GHIJ$.
$$begin{aligned}
A &= frac{p cdot q}{2} \
&= frac{8.94 cdot 4.47}{2} \
&=19.98 text{ units}^2 \
end{aligned}$$

The problem requires determining the length and the width of the garden given the data on its perimeter and a diagonal.

Use the formulas for the perimeter and diagonal of a rectangle to set up a system of equations.

Label the sides of the rectangle with the variables $a$ and $b$.

The perimeter of the rectangle is determined by the formula $P=2a+2b$.

Label the diagonal of the rectangle with the variable $d$. The diagonal of a rectangle bisects it into two right triangles. The diagonal of the rectangle represents the hypotenuse of the right triangles. Use the Pythagorean Principle to determine the formula for the diagonal.

$$begin{aligned}
d^2&=a^2+b^2\
d&=sqrt{a^2+b^2}
end{aligned}$$

Set up the system of equations.
$$
left{
begin{array}{cc}
2a+2b=100 &&text{Equation 1}\
sqrt{a^2+b^2}=40 &&text{Equation 2}
end{array}
right.$$

Transform the system of equations by dividing the entire Equation $1$ by $2$ and squaring the entire Equation $2$.

$$begin{aligned}
a+b&=50 &&text{Equation 3}\
a^2+b^2&=1600&&text{Equation 4}
end{aligned}$$

Isolate the variable $b$ from Equation $3$.

$$begin{aligned}
a+b=50 rightarrow b= 50-a
end{aligned}$$

Insert the obtained expression for the variable $b$ into Equation $4$. Then, solve the equation for the variable $a$.
$$begin{aligned}
a^2+b^2&=1600\
a^2+(50-a)^2&=1600\
a^2+50^2-2cdot50a+a^2&=1600\
2a^2-100a+2500&=1600\
2a^2-100a+900&=0\
a^2-50a+450&=0
end{aligned}$$

Solve the quadratic equation using the Quadratic Formula.
$$begin{aligned}
a=1, b=-50, c=450
end{aligned}$$
$$begin{aligned}
a_{1,2}&= frac{-bpmsqrt{b^2-4ac}}{2a}\
&= frac{-(-50)pmsqrt{(-50)^2-4cdot1cdot450}}{2cdot1}\
&= frac{50pmsqrt{2500-1800}}{2}\
&= frac{50pmsqrt{700}}{2}\
&= frac{50pm10sqrt7}{2}\
a_1&=25+5sqrt7\
&approxboxed{38.23}\
a_2&=25-5sqrt7\
&approxboxed{11.77}
end{aligned}$$

Use the expression $b=50-a$ to calculate the value of the variable $b$.
$$begin{aligned}
b_1&=50-a_1\
&=50-38.23\
&approxboxed{11.77}\
\
b_2&=50-a_2\
&=50-11.77\
&approxboxed{38.23}\
end{aligned}$$

Conclude that the dimensions of the garden are $38.23times 11.77$ meters.

Check your solution. Calculate the perimeter of the garden.
$$begin{aligned}
P&=2a+2b\
&=2cdot38.23+2cdot11.77\
&=76.46+23.54\
&=100
end{aligned}$$

$38.23times 11.77$ meters.

**Solution**
Solve for the volume of the bigger cone.
$$begin{aligned}
V_1 &= frac{pi r^2 h}{3}\
&= frac{pi (6)^2 5}{3} \
&= 188.50 text{ inches}^3
end{aligned}$$

where height became $3$ inches because Matt cut $2$ inches.

Using ratio and proportion, the radius of the smaller cone is $4$ inches.
$$begin{aligned}
V_2 &= frac{pi r^2 h}{3}\
&= frac{pi (4)^2 2}{3} \
&= 33.51 text{ inches}^3
end{aligned}$$

$$begin{aligned}
V &= V_1 – V_2 \
&= 188.50 – 33.51\
&= boxed{154.99 text{ inches}^3} \
end{aligned}$$

**Solution**
Solve the perimeter of the sector.
$$begin{aligned}
P &= 6 text{ ft} + 6 text{ ft} + frac{30^circ}{360^circ} cdot(2pi cdot 6 text{ ft}) \\
&= 12 text{ ft} + frac{30^circ}{360^circ} cdot(12pi text{ ft}) \\
&= 12 text{ ft} + frac{360^circpi}{360^circ}text{ ft}\\
&= 12 + pi text{ ft}\
end{aligned}$$

C. $12 + pi text{ ft}\$

**Solution**
a. The triangles are similar because their angles are congruent.
For $triangle ABC$:
$$begin{aligned}
180^circ &= 43^circ + 20^circ +x \
180 ^circ &= 63^circ +x \
180 ^circ -63^circ &= x \
117 ^circ &= x \
end{aligned}$$

For $triangle EDF$:
$$begin{aligned}
180^circ &= 43^circ + 117^circ +x \
180 ^circ &= 160^circ +x \
180 ^circ -160^circ &= x \
20 ^circ &= x \
end{aligned}$$
Therefore, $triangle ABC$ ~ $triangle EDF$.

b. Compare the sides and notice that the scale factor for all the sides is $1.8$.
$$begin{aligned}
overline{AB} &= 13 times 1.8 = 23.4 = overline{KM}\
overline{BC} &= 10 times 1.8 =18= overline{KL} \
overline{AC} &= 5 times 1.8 =9 = overline{LM}\
end{aligned}$$

Therefore, $triangle ABC$ ~ $triangle KLM$.

c. Compare the sides and notice that there will be no scale factor for the sides.

Therefore, $triangle ABC$ is not similar to $triangle XYZ$.

**Solution**
a. Since the length is $9$ square units and the width is $8$ square units, the actual dimensions are:
$$begin{aligned}
l &= 9cdot 32 \
&= 288 text{ ft} \
\
w &= 8cdot 32 \
&= 256text{ ft} \
end{aligned}$$

b. Compute for the area of the shape.
$$begin{aligned}
A&= l cdot w \
&= 288text{ ft} cdot 256text{ ft}\
&= 73,728 text{ ft}^2
end{aligned}$$

Compute for the area of the triangle in the bottom-left part of the island.
$$begin{aligned}
A&= frac{bcdot h}{2} \
&= frac{74 text{ ft} cdot 74text{ ft}}{2} \
&= 2,738 text{ ft}^2
end{aligned}$$

Compute for the area of the rectangle on the middle right.
$$begin{aligned}
A&= lcdot w\
&= 74 text{ ft} cdot 96text{ ft} \
&= 7,104 text{ ft}^2
end{aligned}$$

Compute for the area of the rectangle on the upper-right part.
$$begin{aligned}
A&= lcdot w\
&= 128 text{ ft} cdot 32 text{ ft} \
&= 4,096 text{ ft}^2
end{aligned}$$

Compute for the area of the triangle in the upper-right part of the island.
$$begin{aligned}
A&= frac{bcdot h}{2} \
&= frac{32 text{ ft} cdot 32text{ ft}}{2} \
&= 512 text{ ft}^2
end{aligned}$$

Add all the other areas and subtract them to the area of the shape to get the actual area of the island.
$$begin{aligned}
&= 73,728 – (2,738+7,104+4,096+512) \
&= 73,728 – 14450 \
&= 59,278 text{ ft}^2\
end{aligned}$$

**Solution**
a. Move the given graph $4$ units to the left.
[](https://postimg.cc/5YLhDbjk)

b. Move the given graph $5$ units below.
[](https://postimg.cc/rR9vFj11)