b. Since 6,8,10 is a Pythagorean triple, we then know that $y=8$ and $y=-8$ are possible $y$-values.

c. Since 3,4,5 is a Pythagorean triple, we then know that $y=4$ and $y=-4$ are possible $y$-values.

d. Using the Pythagorean theorem:

$$
x^2+y^2=10^2=100
$$

e. Yes, it workd for all points that lie a distance of 10 from the center of the origin.

b. $y=8$ and $y=-8$

c. $y=4$ and $y=-4$

d. $x^2+y^2=100$

e. Yes

$$
x^2+y^2=4^2=16
$$

b. Replace the radius 4 with $r$:

$$
x^2+y^2=r^2
$$

c. Yes, it is a circle with radius 6 centered at the origin:

b. $x^2+y^2=r^2$

c. Yes

d. Origin

c. Use the Pythagorean theorem:

$$
(x-3)^2+(y-1)^2=5^2
$$

d. (1)
$$
(x-2)^2+(y-7)^2=1^2
$$

(2)
$$
(x-3)^2+(y-3)^2=2^2
$$

(3) The circle is then centered at the origin with radius 5:

$$
x^2+y^2=5^2
$$

(4)
$$
(x+2)^2+(y-1)^2=3^2
$$

b. $y-1$, $x-3$

c. $(x-3)^2+(y-1)^2=5^2$

d.

(1) $(x-2)^2+(y-7)^2=1^2$

(2) $(x-3)^2+(y-3)^2=2^2$

(3) $x^2+y^2=5^2$

(4) $(x+2)^2+(y-1)^2=3^2$

b. Use the substitution method and replace $y$ with $x+2$ in the second equation:

$$
(x+2)^2+(x+2+3)^2=9
$$

Rewrite the equation:

$$
x^2+4x+4+x^2+10x+25=9
$$

Subtract 9 from both sides of the equation:

$$
x^2+14x+20=0
$$

Determine the roots using the quadratic formula:

$$
x=dfrac{-14pm sqrt{14^2-4(1)(20)}}{2(1)}=-7pm sqrt{29}
$$

Determine $y$:

$$
y=x+2=-7pm sqrt{29}+2=-5pm sqrt{29}
$$

Thus the solutions are $(-7pm sqrt{29},-5pm sqrt{29})$.

c. A line can intersect a circle at most twice and thus we have found all points of intersections.

b. $(-7pm sqrt{29}, -5pm sqrt{29})$

c. A line can intersect a circle at most twice.

The equation of a circle with center $(x_1,y_1)$ and radius $r$ is:

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

$$
7cdot 6=42
$$

b. The bottom surface has 7 surfaces (and the upper surface thus too), while the sides of one layer has 14 surfaces:

$$
2cdot 7+6cdot 14=98
$$

c. (1) Volume 20 and surface area 58

(2) Volume 24 and surface area 60

(3) Volume
$$
3cdot 4cdot 5=60
$$

and surface area
$$
2cdot 3cdot 5+2cdot 3cdot 4+2cdot 4cdot 5=30+24+40=94
$$

b. 98 units$^3$

c. (1) 20 units$^3$ and 58 units$^2$

(2) 24 units$^3$ and 60 units$^2$

(3) 60 units$^3$ and 94 units$^2$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

a. Since the circle has center the origin and radius 3:

$$
x^2+y^2=3^2=9
$$

b. The radius is $sqrt{81}=9$:

Factorize:

$$
y=2(x^2+6x)-7
$$

Add and subtract 18 from both sides of the equation:

$$
y=2(x^2+6x+9)-7-18
$$

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

$$
y=2(x+3)^2-25
$$

Thus the vertex is $(-3,-25)$.

Determine the $x$-intercepts by letting $y=0$:

$$
0=2(x+3)^2-25
$$

Add 25 to both sides of the equation:

$$
25=2(x+3)^2
$$

Divide both sides of the equation by 2:

$$
dfrac{25}{2}=(x+3)^2
$$

Take the square root of both sides of the equation:

$$
pm dfrac{5sqrt{2}}{2}=pm sqrt{dfrac{25}{2}}=x+3
$$

Subtract 3 from both sides of the equation:

$$
-3pm dfrac{5sqrt{2}}{2}=x
$$

Thus the $x$-intercepts are $x=-3pm dfrac{5sqrt{2}}{2}$.

$$
y=2x+3
$$

Interchange $x$ and $y$:

$$
x=2y+3
$$

Subtract 3 from both sides of the equation:

$$
x-3=2y
$$

Divide both sides of the equation by 2:

$$
dfrac{1}{2}(x-3)=y
$$

Interchange the left and right side of the equation and replace $y$ with $f^{-1}(x)$:

$$
f^{-1}(x)=dfrac{1}{2}(x-3)
$$

Thus we note that Kent was not correct.

$$
77cm cdot 53cm=4,081cm^2
$$

The area of the billboard is then the product of the area of the painting and the linear scale factor squared:

$$
20^2cdot 4,081cm^2=1,632,400cm^2
$$

$$
3x-26+2x+70+5x-10+3x+2x+56=540
$$

Combine like terms:

$$
15x+90=540
$$

Subtract 90 from both sides of the equation:

$$
15x=450
$$

Divide both sides of the equation by 15:

$$
x=30
$$

$$
3x-26=3(30)-26=64text{textdegree}
$$

$$
2x+70=2(30)+70=130text{textdegree}
$$

$$
5x-10=5(30)-10=140text{textdegree}
$$

$$
3x=3(30)=90text{textdegree}
$$

$$
2x+56=2(30)+56=116text{textdegree}
$$

Then we note that 4 out of 5 angles is greater than or equal to 90$text{textdegree}$:

$$
P=dfrac{4}{5}=0.8=80%
$$

$$
A=8cdot 9-dfrac{4cdot 4}{2}=72-8=64
$$

$$
h=12sin{79text{textdegree}}approx 11.78
$$

$$
dfrac{b}{2}=12cos{79text{textdegree}}approx 2.29Rightarrow bapprox 4.58
$$

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

$$
A=dfrac{4.58cdot 11.78}{2}approx 27
$$

$$
sqrt{8^2-4^2}=sqrt{48}=4sqrt{3}
$$

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

$$
A=dfrac{4cdot 4sqrt{3}}{2}=8sqrt{3}
$$

b. 27

c. $8sqrt{3}$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

a.
$$
(x-3)^2+(y-1)^2=5^2
$$

b. Use $(apm b)^2=a^2pm 2ab+b^2$:

$$
x^2-6x+9+y^2-2y+1=25
$$

Subtract 25 from both sides of the equation:

$$
x^2+y^2-6x-2y-15=0
$$

c. You can change it back using the method for complete the square (for $x$ and for $y$).

b. $x^2+y^2-6x-2y-15=0$

c. Method for complete the square

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

a. To complete the square containing the $x$ terms needs 4 units, while the square containing the $y$ terms needs 1 unit.

b. Thus we need to add 5 units in total.

c. We need to add 5 units to both sides of the equation.

d. Then we obtain:

$$
(x^2+4x+4)+(y^2+2y+1)=11+5
$$

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

$$
(x+2)^2+(y+1)^2=16=4^2
$$

Thus the equation is a circle with center $(-2,-1)$ and radius 4.

b. Add 5 units in total

c. Add 5 units to both sides of the equation

c. $(x+2)^2+(y+1)^2=16$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

Add 3 to both sides of the given equation:

$$
x^2-4x+y^2-6y=3
$$

Add 13 to both sides of the equation:

$$
(x^2-4x+4)+(y^2-6y+9)=3+13
$$

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

$$
(x-2)^2+(y-3)^2=16=4^2
$$

Thus the equation is a circle with center $(2,3)$ and radius 4.

$$
12x-12=0
$$

Add 12 to both sides of the equation:

$$
12x=12
$$

Divide both sides of the equation by 12:

$$
x=1
$$

Now we need to find the corresponding $y$-values, however this cannot be done directly because the equations contain terms $y^2$ and $y$.

$$
(x+2)^2+(y-3)^2=25
$$

$$
(x-4)^2+(y-3)^2=25
$$

Subtract the two equations:

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

Then we note that $x=1$ is a solution.

Determine the corresponding $y$-value:

$$
(y-3)^2=25-(1-4)^2=25-9=16
$$

Take the square root of both sides of the equatino:

$$
y-3=pm 4
$$

Add 3 to both sides of the equation:

$$
y=3pm 4=7text{ or }-1
$$

Thus the solutions are $(1,7)$ and $(1,-1)$.

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

Subtract 9 from both sides of the given equation:

$$
x^2-8x+y^2+6y=-9
$$

Add 25 to both sides of the equation:

$$
(x^2-8x+16)+(y^2+6y+9)=-9+25
$$

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

$$
(x-4)^2+(y+3)^2=16=4^2
$$

Thus the equation is a circle with center $(4,-3)$ and radius 4.

$$
(x-3)(x+1)<0
$$

Thus the roots are $x=3$ and $x=-1$. Since the function $y=x^2-2x-3$ opens upwards, we then know:

$$
-1<x<3
$$

$$
x^2+4x-3=0
$$

Determine the roots using the quadratic formula:

$$
x=dfrac{-4pm sqrt{4^2-4(1)(-3)}}{2(1)}=-2pm sqrt{7}
$$

$$
(x-4)(x+9)<0
$$

Thus the roots are $x=4$ and $x=-9$. Since the function $y=x^2+5x-36$ opens upwards, we then know:

$$
-9<x<4
$$

$$
x=dfrac{3pm sqrt{(-3)^2-4(1)(13.75)}}{2(1)}=dfrac{3pm sqrt{-46}}{2}=dfrac{3}{2}pm dfrac{sqrt{46}}{2}i
$$

b. $x=-2pm sqrt{7}$

c. $-9<x<4$

d. $x=frac{3}{2}pm frac{sqrt{46}}{2}i$

$$
0=2x^2-x-15
$$

Factorize:

$$
0=(2x+5)(x-3)
$$

Zero product property:

$$
2x+5=0text{ or }x-3=0
$$

Solve each equation to $x$:

$$
2x=-5text{ or }x=3
$$

$$
x=-dfrac{5}{2}text{ or }x=3
$$

b. Same shape, Same $x$-intercept, reflected about the $x$-axis

a.
$$
3+i-(5+6i)=(3-5)+(i-6i)=-2-5i
$$

b.
$$
8i+16+2i=16+(8i+2i)=16+10i
$$

c.
$$
(20+5i)-(-5i)=20+(5i-(-5i))=20+10i
$$

b. $16+10i$

c. $20+10i$

$$
V=16^3=4,096units^3
$$

The surface area contains six squares:

$$
S=6cdot 16^2=1,536units^2
$$

Surface area 1;536 units$^2$

$$
P(2blue)=P(blue)cdot P(blue)=dfrac{5}{6}cdot dfrac{5}{6}=dfrac{25}{36}approx 0.694=69.4%
$$

c. The area of a circle sector with central angle $theta$ is $dfrac{theta}{360text{textdegree}}pi r^2$:

$$
A=dfrac{300text{textdegree}}{360text{textdegree}}pi 7^2=dfrac{245pi}{6} cm^2
$$

d. The central angle of the purple area is 90$text{textdegree}$ and the area of the yellow area is 240$text{textdegree}$. Since the total spinner is 360$text{textdegree}$, the central angle of the green region is then:

$$
360text{textdegree}-240text{textdegree}-90text{textdegree}=30text{textdegree}
$$

b. $frac{25}{36}approx 0.694=69.4%$

c. $frac{245 pi}{6}$ cm$^2$

d. 30$text{textdegree}$

$$
4x^2-12x=4x(x-3)
$$

$$
3y^2+6y+3=3(y^2+2y+1)
$$

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

$$
3y^2+6y+3=3(y+1)^2
$$

$$
2m^3+7m^2+3m=m(2m^2+7m+3)
$$

Factor further:

$$
2m^3+7m^2+3m=m(2m+1)(m+3)
$$

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

b. $3(y+1)^2$

c. $m(2m+1)(m+3)$

d. $(3x-2)(x+2)$

b. Yes, The curve (parabola) lies with vertex between F and the line and opens away from the line $l$.

c. No, there is only one point.

d.

b. $R$ is three circles from the focus

c. No

d. Sketch

$$
sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=sqrt{(x-0)^2+(y-0)^2}=sqrt{x^2+y^2}
$$

c. $y=-2$

d. The distance is the difference in $y$-coordinates: $y+2$.

e. Since every point is an equal distance from the focus and the directrix:

$$
sqrt{x^2+y^2}=y+2
$$

Square both sides of the equation:

$$
x^2+y^2=y^2+4y+4
$$

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

$$
x^2=4y+4
$$

Subtract 4 from both sides of the equation:

$$
x^2-4=4y
$$

Divide both sides of the equation by:

$$
dfrac{1}{4}x^2-1=y
$$

Thus the equation of the parabola is $y=dfrac{1}{4}x^2-1$.

b. $sqrt{x^2+y^2}$

c. $y=-2$

d. $y+2$

e. $y=frac{1}{4}x^2-1$

$$
sqrt{(x+3)^2+(y+4)^2}=y+2
$$

Square both sides of the equation:

$$
(x+3)^2+(y+4)^2=y^2+4y+4
$$

Expand:

$$
x^2+6x+9+y^2+8y+16=y^2+4y+4
$$

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

$$
x^2+6x+25+8y=4y+4
$$

Subtract $8y$ from both sides of the equation:

$$
x^2+6x+25=-4y+4
$$

Subtract 4 from both sides of the equation:

$$
x^2+6x+21=-4y
$$

Divide both sides of the equation by $-4$:

$$
-dfrac{1}{4}(x^2+6x+21)=y
$$

Thus the equation of the parabola is $y=-dfrac{1}{4}(x^2+6x+21)$.

$$
y=-dfrac{1}{4}(x^2+6x+9)-dfrac{21}{4}+dfrac{9}{4}
$$

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

$$
y=-dfrac{1}{4}(x+3)^2-3
$$

b. $y=-frac{1}{4}(x^2+6x+21)$

c. $y=-frac{1}{4}(x+3)^2-3$

$$
8x+20-11=4x-3
$$

Subtract $4x$ from both sides of the equation:

$$
4x+9=-3
$$

Subtract 9 from both sides of the equation:

$$
4x=-12
$$

Divide both sides of the equation by 4:

$$
x=-3
$$

$$
20m-10=19m
$$

Subtract $20m$ of both sides of the equation:

$$
-10=-m
$$

Divide both sides of the equation by $-1$:

$$
10=m
$$

$$
(3p-2)(p+4)=0
$$

Zero product property:

$$
3p-2=0text{ or }p+4=0
$$

Solve each equation to $p$:

$$
3p=2text{ or }p=-4
$$

$$
p=dfrac{2}{3}text{ or }p=-4
$$

$$
x+2=25
$$

Subtract 2 from both sides of the equation:

$$
x=23
$$

b. $m=10$

c. $p=frac{2}{3}$ or $p=-4$

d. $x=23$

Let $x$ be the time in years and $y$ the total amount of money.

Then we obtain:

$$
f(x)=left{ begin{matrix} 200 & 0leq xleq 0.25\ 200(1+0.02)^{x-0.25} & 0.25<xleq 5.25end{matrix}right.
$$

$$
15.4in^2=dfrac{15.4}{12^2}ft^2approx 0.107ft^2
$$

Thus you get more pizza from the square pizza and thus you should order the slice of the square pizza.

$$
dfrac{360text{textdegree}}{5}=72text{textdegree}
$$

Thus we then also know that the two base angles of the isosceles triangles are 72$text{textdegree}$. Since the sum of all angles in a triangle is 180$text{textdegree}$, we then know that the vertex angles are:

$$
180text{textdegree}-72text{textdegree}-72text{textdegree}=36text{textdegree}
$$

b. The first model seems the best, because the function keep decreasing but will not become negative and will not increase again.

c. Replace $x$ with 10, 20 and 30:

$$
y=16+10(0.95)^{10}approx 22
$$

$$
y=16+10(0.95)^{20}approx 20
$$

$$
y=16+10(0.95)^{30}approx 18
$$

d. The distance that the athletes run, does not determine the intensity of the training and thus the plan will not work for everybody.

a.
$$
(x+i)(x-i)=x^2+ix-ix-i^2=x^2-i^2=x^2+1
$$

b.
$$
(x+2i)(x-2i)=x^2+2xi-2xi-4i^2=x^2-4i^2=x^2+4
$$

c. Use the difference of squares:

$$
x^2+25=x^2-(-25)=x^2-(25i^2)=x^2-(5i)^2=(x-5i)(x+5i)
$$

b. $x^2+4$

c. $(x-5)(x+5i)$

Multiply both sides of the equation by 10:

$$
5=BC
$$

$$
cos{B}=sin{A}=0.5
$$

$$
AC=sqrt{10^2-5^2}=sqrt{75}=5sqrt{3}
$$

b. 5

c. 0.5

d. $5sqrt{3}$

$$
22=51-29<x<51+29=80
$$

the length needs to be more than 22 and less than 80, which means that (C) is the only possible answer.

$$
sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=sqrt{(x-0)^2+(y-0)^2}=sqrt{x^2+y^2}
$$

b. The distance is the difference in $y$-coordinates: $y+6$.

c. Since every point is an equal distance from the focus and the directrix:

$$
sqrt{x^2+y^2}=y+6
$$

Square both sides of the equation:

$$
x^2+y^2=y^2+12y+36
$$

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

$$
x^2=12y+36
$$

Subtract 36 from both sides of the equation:

$$
x^2-36=12y
$$

Divide both sides of the equation by:

$$
dfrac{1}{12}x^2-3=y
$$

Thus the equation of the parabola is $y=dfrac{1}{12}x^2-3$.

b. $y+6$

c. $y=frac{1}{12}x^2-3$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

a. The center is $(2,-3)$ and the radius is 5.

b. Add 6 to both sides of the given equation:

$$
x^2+2x+y^2+6y=6
$$

Add 10 to both sides of the equation:

$$
(x^2+2x+1)+(y^2+6y+9)=6+10
$$

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

$$
(x+1)^2+(y+3)^2=16=4^2
$$

Thus the equation is a circle with center $(-1,-3)$ and radius 4.

b. $(-1,-3)$, 4

$$
y=dfrac{x}{3}-2
$$

Interchange $x$ and $y$:

$$
x=dfrac{y}{3}-2
$$

Add 2 to both sides of the equation:

$$
x+2=dfrac{y}{3}
$$

Multiply both sides of the equation by 3:

$$
3(x+2)=y
$$

Interchange the left and right side of the equation and replace $y$ with $f^{-1}(x)$:

$$
f^{-1}(x)=3(x+2)=3x+6
$$

$$
y=dfrac{1}{2}x+5
$$

Interchange $x$ and $y$:

$$
x=dfrac{1}{2}y+5
$$

Subtract 5 from both sides of the equation:

$$
x-5=dfrac{1}{2}y
$$

Multiply both sides of the equation by 2:

$$
2(x-5)=y
$$

Interchange the left and right side of the equation and replace $y$ with $g^{-1}(x)$:

$$
g^{-1}(x)=2(x-5)=2x-10
$$

b. $g^{-1}(x)=2x-10$

Since the area is double every 20 minutes, the area is multiplied by $2^6=64$ in 2 hours (as there are six 20-minute periods in 2 hours).

b.
$$
1(2)^{-0.5}
$$

$$
1(2^{1/20})^{-10}
$$

$$
1(2^3)^{1/6}
$$

b. $1(2)^{-0.5}$

b. Let $g(x)$ be 0:

$$
0=x^2+9
$$

Subtract 9 from both sides of the equation:

$$
-9=x^2
$$

Take the square root of both sides of the equation:

$$
pm 3i = pm sqrt{-9}=x
$$

Thus the roots are $x=pm 3i$ and we can then write:

$$
x^2+9=(x-3i)(x+3i)
$$

b. $x^2+9=(x-3i)(x+3i)$

$$
x-3=dfrac{11}{4}
$$

Add 3 to both sides of the equation:

$$
x=dfrac{11}{4}+3=dfrac{23}{4}
$$

$$
x=pm sqrt{-10}=pm sqrt{10}i
$$

$$
3x^2=18
$$

Divide both sides of the equation by 3:

$$
x^2=6
$$

Take the square root of both sides of the equation:

$$
x=pm sqrt{6}
$$

b. $x=pm sqrt{10}i$

c. $x=pm sqrt{6}$

d. No solutions

$$
a=dfrac{360text{textdegree}}{3}=120text{textdegree}
$$

while we know that the interior angle of a regular pentagon is $b=108text{textdegree}$. Thus $a$ is greater.

$$
a=sin^{-1}dfrac{8}{12}approx 42text{textdegree}
$$

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

$$
b=cos^{-1}dfrac{8}{12}approx 48text{textdegree}
$$

Thus $b$ is greater.

b. Not enough information

c. $b$

d. $a$

e. Same

$$
dfrac{360text{textdegree}}{18text{textdegree}}=20
$$

$$
dfrac{360text{textdegree}}{24text{textdegree}}=15
$$

$$
dfrac{360text{textdegree}}{28text{textdegree}}approx 12.857
$$

$$
dfrac{360text{textdegree}}{40text{textdegree}}=9
$$

Thus we note that C cannot be the measure of an exterior angle of a regular polygon.

$$
1.85cdot 10=18.5ft
$$

The diameter is twice the radius:

$$
2cdot 18.5ft=37ft
$$

Thus the diameter dos not appear larger than 45 ft.

b. Let us call the larger arc $overset{frown}{AXB}$.

b. Let us call the larger arc $overset{frown}{AXB}$.

c. If the minor and major arc have the same length, then the points $A$ and $B$ lie on direct opposite sides of the circle and thus the chord is then the diameter of the circle.

$$
overline{AD}cong overline{BD}(text{Given})
$$

$$
overline{AC}cong overline{BC}(text{Radius circle})
$$

$$
overline{CD}cong overline{CD}(text{Common side})
$$

b. However then we know that $mangle ADC=mangle BDC$. Since $C$ does not lie in the center we know that at least one of these angle is less than 90$text{textdegree}$ and thus
$$
mangle ADC+mangle BDC<180text{textdegree}
$$

But this then means that $overline{AB}$ cannot be a chord of the circle which is a contradiction. Thus the center must lie on th perpendicular bisector.

b. $mangle ADC+mangle BDC<180text{textdegree}$

$$
x=sqrt{5^2-4^2}=sqrt{9}=3
$$

Thus the chord lies 3 units away from the center.

$$
mangle ABC=180text{textdegree}-35text{textdegree}-35text{textdegree}=110text{textdegree}
$$

$angle ABC$ and $angle CBD$ are supplementary angles:

$$
mangle CBD=180text{textdegree}-mangle ABC=180text{textdegree}-110text{textdegree}=70text{textdegree}
$$

$$
mangle ABC=180text{textdegree}-mangle CBD=180text{textdegree}-100text{textdegree}=80text{textdegree}
$$

Triangle $ABC$ is isosceles since two of its sides are the radius of the circle. Thus the base angles of the triangle are congruent and the sum of all angles in a triangle is 180$text{textdegree}$:

$$
mangle A=dfrac{180text{textdegree}-80text{textdegree}}{2}=50text{textdegree}
$$

$$
mangle CBD=180text{textdegree}-x-x=180text{textdegree}-2x
$$

$angle ABC$ and $angle CBD$ are supplementary angles:

$$
mangle CBD=180text{textdegree}-mangle ABC=180text{textdegree}-(180text{textdegree}-2x)=2x
$$

(b) 50$text{textdegree}$

(c) $2x$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

a. The circle has center $(0,0)$ and radius 4.5:

$$
(x^2-4x+4)+(y^2-2y+1)=15+4
$$

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

$$
(x-2)^2+(y-1)^2=19
$$

Thus the equation is a circle with center $(2,1)$ and radius $sqrt{19}approx 4.36$.

b. $(0,0)$, $sqrt{75}$

c. $(3,0)$, 1

d. $(2,1)$, $sqrt{19}$

$$
dfrac{312}{13}=24
$$

Thus the tower is 24 layers tall.

$$
x^2+dfrac{9}{16}x^2=25
$$

Combine like terms:

$$
dfrac{25}{16}x^2=25
$$

Multiply both sides of the equation by $dfrac{16}{25}$:

$$
x^2=16
$$

Take the square root of both sides of the equation:

$$
x=pm 4
$$

Determine $y$

$$
y=dfrac{3}{4}(pm 4)=pm 3
$$

Thus the solutions are then $(pm 4,pm 3)$

b. The average rate of change is the difference in value divided by the difference in months:

$$
A=dfrac{1060-1040}{6-4}=10
$$

$$
B=dfrac{1061.50-1040.60}{6-4}=10.45
$$

$$
C=dfrac{1036-1016}{6-4}=10
$$

c. Investment B, because in the long run the exponential always wins over the linear and quadratic.

Add 2 to both sides of the equation:

$$
dfrac{4}{5}x=9
$$

Multiply both sides of the equation by $dfrac{5}{4}$:

$$
x=dfrac{35}{4}
$$

Divide both sides of the equation by 3:

$$
x^2=100
$$

Take the square root of both sides of the equation:

$$
x=pm 10
$$

$$
12x-3=2x+10
$$

Subtract $2x$ from both sides of the equation

$$
10x-3=10
$$

Add 3 to both sides of the equation:

$$
10x=13
$$

Divide both sides of the equatino by 10:

$$
x=dfrac{13}{10}=1.3
$$

$$
x=dfrac{4pm sqrt{(-4)^2-4(1)(6)}}{2(1)}=2pm sqrt{2}i
$$

b. $x=pm 10$

c. $x=frac{13}{10}=1.3$

d. $s=2pm sqrt{2}i$

$$
x(3x-2)=481
$$

The perimeter is the sum of all sides:

$$
2x+2(3x-2)=100
$$

c. Rewrite the equations:

$$
3x^2-2x-481=0
$$

$$
8x-4=100
$$

Solve the second equation. Add 4 to both sides of the equation:

$$
8x=104
$$

Divide both sides of the equation by 8:

$$
x=13
$$

Check if it is also a solution of the first equation:

$$
3(13)^2-2(13)-481=0
$$

Thus we then know that the width is 13ft and the length is:

$$
l=3x-2=3(13)-2=37ft
$$

b. $x(3x-2)=481$, $2x+2(3x-2)=100$

c. 13 ft by 37 ft

$$
Downarrow
$$

$$
overline{KM}=overline{MQ}
$$

$$
angle KML cong angle QMP(text{Vertical angles})
$$

$$
angle L cong angle P(text{Given})
$$

$$
Downarrow AAS
$$

$$
triangle KML cong triangle QMP
$$

$$
Downarrow
$$

$$
overline{KL}cong overline{QP}
$$

$$
theta=tan^{-1}dfrac{127ml}{1000mm}approx 7text{textdegree}
$$

b. Since alternate intior angles of parallel lines are congruent, we then know that the central angle is about $7text{textdegree}$

c. The circumference of $theta=7text{textdegree}$ is 500 and thus the circumference of the earth (which is 360$text{textdegree}$) is then:

$$
dfrac{360text{textdegree}}{7text{textdegree}}cdot 500approx 25,714mi^2
$$

d. Since the circumference is also $C=2pi r$:

$$
r=dfrac{C}{2pi}=dfrac{25,714}{2pi}approx 4,093mi
$$

b. 7$text{textdegree}$

c. 25,714 mi$^2$

d. 4,903 mi

$$
moverset{frown}{SU}=40text{textdegree}
$$

The major arc is then:

$$
moverset{frown}{SZU}=360text{textdegree}-40text{textdegree}=320text{textdegree}
$$

b. The arce length is the product of the measure of the arc to the entire circle and the circumference of a circle($C=2pi r$):

$$
dfrac{40text{textdegree}}{360text{textdegree}}cdot (2pi (12))=dfrac{8}{3}piapprox 7.11
$$

b. $frac{8}{3}pi approx 7.11$

$$
mangle F=mangle G=mangle H
$$

b. Both angles intercept the same arc $overset{frown}{WY}$ and we note that the measure of the central angle is double the measure of the inscribed angle.

$$
x=dfrac{118text{textdegree}}{2}=59text{textdegree}
$$

$$
x=2cdot 41text{textdegree}=82text{textdegree}
$$

$$
x=180text{textdegree} (=2cdot 90text{textdegree})
$$

$$
y=2cdot 78text{textdegree}=156text{textdegree}
$$

The sum of all angles in a triangle is also equal to 180$text{textdegree}$:

$$
z=2cdot (180text{textdegree}-78text{textdegree}-90text{textdegree})=2cdot 12text{textdegree}=24text{textdegree}
$$

$$
x=56text{textdegree}
$$

The inscribed angle is half the central angle:

$$
y= dfrac{56text{textdegree}}{2}=28text{textdegree}
$$

$$
x=114text{textdegree}
$$

The inscribed angle is half the central angle:

$$
y= dfrac{114text{textdegree}}{2}=57text{textdegree}
$$

$$
x=2cdot 28text{textdegree}=56text{textdegree}
$$

The inscribed angles of the same arc is are congruent:

$$
y= 56text{textdegree}
$$

b. 82$text{textdegree}$

c. 180$text{textdegree}$, 156$text{textdegree}$, 24$text{textdegree}$

d. 56$text{textdegree}$, 28$text{textdegree}$

e. 114$text{textdegree}$, 57$text{textdegree}$

f. 56$text{textdegree}$, 56$text{textdegree}$

$$
mangle D=mangle C=64text{textdegree}
$$

$$
moverset{frown}{BF}=2mangle C=2cdot 64text{textdegree}=128text{textdegree}
$$

$$
mangle E=mangle C=64text{textdegree}
$$

$$
moverset{frown}{CBF}=180text{textdegree}
$$

$$
mangle BAF=2mangle BCF=2cdot 64text{textdegree}=128text{textdegree}
$$

$$
mangle BAC = 180text{textdegree}-mangle BAF=180text{textdegree}-128text{textdegree}=52text{textdegree}
$$

b. 128$text{textdegree}$

c. 64$text{textdegree}$

d. 180$text{textdegree}$

e. 128$text{textdegree}$

f. 52$text{textdegree}$

$$
mangle EDC = dfrac{5-2}{5}180text{textdegree}=108text{textdegree}
$$

$$
mangle BOC=180text{textdegree}-108text{textdegree}=72text{textdegree}
$$

$$
moverset{frown}{EBC}=2cdot 72text{textdegree}=144text{textdegree}
$$

b. 108$text{textdegree}$

c. 72$text{textdegree}$

d. 144$text{textdegree}$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

a. The circle has center $(-5,0)$ and radius $sqrt{10}approx 3.16$

$$
x^2-6x+y^2-2y=5
$$

Add 10 to both sides of the equation:

$$
(x^2-6x+9)+(y^2-2y+1)=5+10
$$

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

$$
(x-3)^2+(y-1)^2=15
$$

Thus the equation is a circle with center $(3,1)$ and radius $sqrt{15}approx 3.87$.

b. $(3,1)$, $sqrt{15}approx 3.87$

$$
16-2x=3x
$$

Add $2x$ to both sides of the equation:

$$
16=5x
$$

Divide both sides of the equation by 5:

$$
3.2=dfrac{16}{5}=x
$$

$$
-10x+2-3=-10x
$$

Add $10x$ to both sides of the equation:

$$
-1=0
$$

Thus is impossible because $-1$ does not equal 0 and thus the equation has no solutions.

$$
(x+11)(x-3)=0
$$

Zero product property:

$$
x+11=0text{ or }x-3=0
$$

Solve each equation to $x$:

$$
x=-11text{ or }x=3
$$

Multiply both sides of the equation by 3:

$$
2x-36=540
$$

Add 36 to both sides of the equation:

$$
2x=576
$$

Divide both sides of the equation by 2:

$$
x=288
$$

b. No solution

c. $x=-11$ or $x=3$

d. $x=288$

$$
y=7x-2
$$

Interchange $x$ and $y$:

$$
x=7y-2
$$

Add 2 to both sides of the equation:

$$
x+2=7y
$$

Divide both sides of the equation by 7:

$$
dfrac{1}{7}(x+2)=y
$$

Interchange the left and right side of the equation and replace $y$ with $f^{-1}(x)$:

$$
f^{-1}(x)=dfrac{1}{7}(x+2)
$$

$$
f^{-1}(x)=dfrac{1}{7}(5+2)=dfrac{1}{7}(7)=1
$$

b. $f^{-1}(x)=1$

$$
sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=sqrt{(x-0)^2+(y-0)^2}=sqrt{x^2+y^2}
$$

b. The distance is the difference in $y$-coordinates: $y-4$.

c. Since every point is an equal distance from the focus and the directrix:

$$
sqrt{x^2+y^2}=y-4
$$

Square both sides of the equation:

$$
x^2+y^2=y^2-8y+16
$$

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

$$
x^2=-8y+16
$$

Subtract 16 from both sides of the equation:

$$
x^2-16=-8y
$$

Divide both sides of the equation by $-8$:

$$
-dfrac{1}{8}x^2+2=y
$$

Thus the equation of the parabola is $y=-dfrac{1}{8}x^2+2$.

b. $y-4$

c. $y=-frac{1}{8}x^2+2$

The area is the product of the length and the width:

$$
9cdot 15=135units^2
$$

The perimeter is the sum of all sides:

$$
9+15+9+15=48units
$$

REDUCED RECTANGLE

The area is the product of the area of the original rectangle and the scale factor squared:

$$
135cdot dfrac{1}{3^2}=15units^2
$$

The perimeter is the product of the perimeter of the original rectangle and the scale factor:

$$
48cdot dfrac{1}{3}=16units
$$

Reduced rectangle: 15 units$^2$, 16 units

$$
C(x)=300+7x
$$

b. Let $P(x)$ be the profit (which is also the revenue decreased by the cost):

$$
P(x)=50x-(300+7x)=43x-300
$$

c.
$$
P(x)=100
$$

Replace the function with its expression:

$$
43x-300=100
$$

Add 300 to both sides of the equation:

$$
43x=400
$$

Divide both sides of the equation by 43:

$$
x=dfrac{400}{43}approx 9.3
$$

Thus she will require at least 10 guests to make a profit of at least $100.

b. $P(x)=43x-300$

c. 10 guests

$$
mangle P=dfrac{180text{textdegree}}{2}=90text{textdegree}
$$

b. $overline{QR}$ would be the diameter of the circle, since $angle QCR$ (with $C$ the center of the circle) is 180$text{textdegree}$.

c. No, because it is still an inscribed angle on the same arc.

Determine the diameter using the Pythagorean theorem:

$$
text{Diameter}=sqrt{8^2+6^2}=sqrt{100}=10
$$

The radius is half the diameter:

$$
begin{align*}
text{Radius}&=frac{text{Diameter}}{2}=frac{10}{2}=5
end{align*}
$$

The area of a circle is the product of $pi$ and the square radius. The radiis half the diameter.

$$
A=pi r^2=pi 5^2=25pi
$$

b. Diameter circle

c. No

d. 25$pi$ units$^2$

$$
moverset{frown}{BAD}=2cdot 100text{textdegree}=200text{textdegree}
$$

The sum of the minor and major arc is 360$text{textdegree}$:

$$
moverset{frown}{BCD}=360text{textdegree}-moverset{frown}{BAD}=360text{textdegree}-200text{textdegree}=160text{textdegree}
$$

The insribed angle of the arc is half the measure of the arc:

$$
x=dfrac{160text{textdegree}}{2}=80text{textdegree}
$$

$$
moverset{frown}{BAD}=2y
$$

The sum of the minor and major arc is 360$text{textdegree}$:

$$
moverset{frown}{BCD}=360text{textdegree}-moverset{frown}{BAD}=360text{textdegree}-2y
$$

The insribed angle of the arc is half the measure of the arc:

$$
x=dfrac{360text{textdegree}-2y}{2}=180text{textdegree}-y
$$

Thus $x$ and $y$ are always supplementary.

b. Yes

c. They have to be similar, because they have two pair of congruent angles (AA).

d. Corresponding sides of similar triangles have the same proportions:

$$
dfrac{DE}{EB}=dfrac{AE}{EC}
$$

Replace the sides with their known lengths:

$$
dfrac{8}{6}=dfrac{4}{EC}
$$

Use cross multiplication:

$$
8EC=4(6)
$$

Divide both sides of the equation by 8:

$$
EC=dfrac{24}{8}=3
$$

$$
sin{75text{textdegree}}=dfrac{x}{6}
$$

Multiply both sides of the equation by 6:

$$
5.8approx 6sin{75text{textdegree}}=x
$$

The length of the chord is then:

$$
LM=2cdot 5.8=11.6units
$$

$$
x=sin^{-1}{dfrac{3}{5}}approx 37text{textdegree}
$$

The length of the arc is then:

$$
moverset{frown}{LM}=2cdot 37text{textdegree}=74text{textdegree}
$$

b. 74$text{textdegree}$

$$
dfrac{x}{6}=dfrac{2}{3}
$$

Multiply both sides of the equation by 6:

$$
x=dfrac{2cdot 6}{3}=4
$$

$$
moverset{frown}{RT}=mangle ROT=180text{textdegree}
$$

Since $S$ is the inscribed angle on $overset{frown}{RT}$:

$$
mangle S=90text{textdegree}
$$

The cosine is the adjacent rectangular side divided by the hypotenuse:

$$
x=cos^{-1}dfrac{r}{2r}=cos^{-1}dfrac{1}{2}=60text{textdegree}
$$

$$
moverset{frown}{EF}=2cdot 52text{textdegree}=104text{textdegree}
$$

The arc of the total circle is 360$text{textdegree}$:

$$
x=moverset{frown}{FG}=360text{textdegree}-104text{textdegree}-131text{textdegree}=125text{textdegree}
$$

b. 60$text{textdegree}$

c. 125$text{textdegree}$

The measure of an arc of which the diameter forms the central angle is 180$text{textdegree}$ and the inscribed angle on the arc is then 90$text{textdegree}$.

Opposite angles of a quadrilateral inscribed in a circle are supplementary.

Two chords of a circle that intersect, create two similar triangles.

You can determine the length of the chord, radius, distance to the center or the central angle using the sine/cosine/tangent ratios.

Opposite angles of a quadrilateral inscribed in a circle are supplementary.

Two chords of a circle that intersect, create two similar triangles.

You can determine the length of the chord, radius, distance to the center or the central angle using the sine/cosine/tangent ratios.

$$
moverset{frown} instead of overarc{AD}=2x=56text{textdegree}
$$

b. Since $overline{AB}$ is the diameter, we know that $mangle D=90text{textdegree}$. Then we can determine $AB$ using the Pythagorean theorem:

$$
AB=sqrt{5^2+(5sqrt{3})^2}=sqrt{100}=10
$$

Then we know that the radius is 5. The area of a circle is $pi$ multiplied by the radius squared:

$$
A=pi r^2=pi 5^2=25pi
$$

c. Since $overline{AB}$ is the diameter, we know that $mangle D=90text{textdegree}$. Since the inscribed angle on an arc is half the measure of the arc:

$$
x=dfrac{100text{textdegree}}{2}=50text{textdegree}
$$

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

$$
BD=2(8)sin{50text{textdegree}}approx 12.26
$$

b. 25$pi$

c. 12.26

$$
6cdot 4cdot 5+6cdot 10cdot 3=120+180=300cm^3
$$

$$
x=y
$$

b. The central angles is double the inscribed angle on the same arc:

$$
y=2x
$$

c. The ratios of intersecting chords are proportional:

$$
dfrac{3}{x}=dfrac{5}{y}
$$

d. Opposite angles of a quadrilateral inscribed in a circle are supplementary:

$$
x+y=180text{textdegree}
$$

b. $y=2x$

c. $frac{3}{x}=frac{5}{y}$

d. $x+y=180text{textdegree}$

$$
16x^2-16x-96leq 0
$$

Factorize:

$$
16(x-3)(x+2)leq 0
$$

Since the function $y=16(x-3)(x+2)$ opens upwards and since $x=3$ and $x=-2$ are its roots:

$$
-2leq xleq 3
$$

$$
x-1=pm 3
$$

Add 1 to both sides of the equation:

$$
x=1pm 3=4text{ or }-2
$$

$$
(x+5)(x-4)<0
$$

Since the function $y=(x+5)(x-4)$ opens upwards and since $x=-5$ and $x=4$ are its roots:

$$
-5<x<4
$$

$$
2(x^2-3x)=-5
$$

Add 4.5 to both sides of the equation:

$$
2(x^2-3x+2.25)=-0.5
$$

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

$$
2(x-1.5)^2=-0.5
$$

Divide both sides of the equation by 2:

$$
(x-1.5)^2=-0.25
$$

Take the square root of both sides of the equation:

$$
x-1.5=pm sqrt{-0.25}=pm 0.5i
$$

Add 1.5 to both sides of the equation:

$$
x=1.5pm 0.5i
$$

b. $x=4$ and $x=-2$

c. $-5<x<4$

d. $x=1.5pm 0.5i$

$$
0=x^2-2x+4
$$

Subtract 3 from both sides of the equation:

$$
-3=x^2-2x+1
$$

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

$$
-3=(x-1)^2
$$

Take the square root of both sides of the equation:

$$
sqrt{3}i=sqrt{-3}=x-1
$$

Add 1 to both sides of the equation:

$$
1pm sqrt{3}i=x
$$

Determine $y$:

$$
y=x-5=1pm sqrt{3}i-5=-4pm sqrt{3}i
$$

Thus the solutions are $(1pm sqrt{3}i, -4pm sqrt{3}i)$

$$
0=2x^2+2x-4
$$

Factorize:

$$
0=2(x+2)(x-1)
$$

Zero product property

$$
x+2=0text{ or }x-1=0
$$

Solve each equation to $x$:

$$
x=-2text{ or }x=1
$$

Determine $y$:

$$
y=x-5=-2-5=-7
$$

$$
y=x-5=1-5=-4
$$

Thus the solutions are $(-2,-7)$ and $(1,-4)$.

b. $(-2,-7)$ and $(1,-4)$

c. The maximum value is the $y$-value of the vertex and thus is $y=1$.

b. $x<-4$

c. $y=1$

c. Yes, because the base angles of an isosceles triangle are congruent. But since the sum of all angles in a triangle is 180$text{textdegree}$, we then know that this is impossible.

d. This means that $overline{AB}$ cannot be a secant line of the circle.

b. Yes

c. Yes

d. $overline{AB}$ cannot be a secant line of the circle

c. At the center of the circle and thus at $(2,3)$. Because the center of the circle is an equal distance away from all three cities.

b. 5, Circumcenter

c. $(2,3)$

$$
angle EAPcong angle DAP (text{$overline{AG}$ is the bisector of $angle BAC$})
$$

$$
angle PEAcong angle PDA (text{Right angles})
$$

$$
Downarrow AAS
$$

$$
triangle PEA cong triangle PDA
$$

$$
Downarrow
$$

$$
PE=PD
$$

$$
angle PBDcong angle PBF(text{$overline{BH}$ is the bisector of $angle ABC$})
$$

$$
angle PDBcong angle PFB(text{Right angles})
$$

$$
Downarrow AAS
$$

$$
triangle PDBcong triangle PFB
$$

$$
Downarrow
$$

$$
PF=PD
$$

Each side of the triangle is then also tangent, because it touches the circle in exactly one point and we know that $overline{PD}$ is perpendicular to $overline{AB}$ (similar to the other sides).

$$
CP=sqrt{6^2-5^2}=sqrt{11}approx 3.32
$$

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

$$
mangle C=sin^{-1}dfrac{5}{6}=56text{textdegree}
$$

$$
moverset{frown}{GD}=360text{textdegree}-40text{textdegree}-210text{textdegree}=110text{textdegree}
$$

$$
moverset{frown}{OR}=360text{textdegree}-40text{textdegree}-110text{textdegree}-180text{textdegree}=30text{textdegree}
$$

$$
moverset{frown}{RGO}=360text{textdegree}-30text{textdegree}=330text{textdegree}
$$

$$
BC=sqrt{34^2-16^2}=sqrt{900}=30
$$

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

$$
moverset{frown}{BC}=2mangle A=2sin^{-1}dfrac{30}{34}approx 124text{textdegree}
$$

b. 330$text{textdegree}$

c. 124$text{textdegree}$

$$
ER=AB=sqrt{39^2-6^2}=sqrt{1485}=3sqrt{165}approx 38.5
$$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

Thus the given circle is a circle with center $(0,3)$ and radius 5.

The $x$-intercepts are the intersections with the $x$-axis and thus are $x=pm 4$. The $y$-intercepts are the intersections with the $y$-axis and thus are $y=-2$ and $y=8$.

$y$-intercepts: $y=-2$ and $y=8$

$$
dfrac{AC}{2}=10sin{40text{textdegree}}approx 6.43
$$

Multiply both sides of the equation by 2:

$$
ACapprox 12.86
$$

$$
dfrac{GK}{3}=dfrac{6}{9}
$$

Multiply both sides of the equation by $3$:

$$
GK=dfrac{6cdot 3}{9}=2
$$

$$
dfrac{360text{textdegree}}{9}=40text{textdegree}
$$

b. Inscribed angles on the same arc are congruent:

$$
mangle C = mangle B=97text{textdegree}
$$

The measure of the arc is double the measure of the inscribed angle on the same arc:

$$
moverset{frown}{AD}=2cdot 97text{textdegree}=194text{textdegree}
$$

c. The measure of the arc is equal to the measure of the central angle over the same arc:

$$
moverset{frown}{AB}=125text{textdegree}
$$

The length of the chord is the ratio of the measure of the arc to that of the circle multiplied by the circumference of a circle ($2pi r$):

$$
dfrac{125text{textdegree}}{360text{textdegree}}cdot 2pi 8=5.56piapprox 17.45
$$

b. 194$text{textdegree}$

c. 17.45

$$
x+3=sqrt{26^2-10^2}=sqrt{576}=24
$$

Subtract 3 from both sides of the equation:

$$
x=21
$$

b. If $x=0$, then $y=-2$ and $a=1$ thus the function opens downwards, then graph 3 is the only possible graph.

c. If $x=0$, then $y=2$, then graph 1 is the only possible graph.

$$
V=pi 3^2cdot 2=18pi
$$

The volume of the remaining piece is then the product of the found volume and the ratio of the angle to the angle of a full circle:

$$
V=dfrac{45text{textdegree}}{360text{textdegree}}cdot 18pi=dfrac{9}{4}pi in^3approx 7in^3
$$

$$
2x+3x+4x+5x=360text{textdegree}
$$

Combine like terms:

$$
14x=360text{textdegree}
$$

Divide both sides of the equation by 14:

$$
x=dfrac{180}{7}text{textdegree}approx 26text{textdegree}
$$

where $overset{largefrown}{DF}$ is the arc angle from point $D$ and $F$.

$$
angle B=dfrac{180text{textdegree}}{2}=90text{textdegree}
$$

c.

$$
angle B=dfrac{180text{textdegree}}{2}=90text{textdegree}
$$

When $angle B$ is a right angle, we thus obtain that $overline{AB}$ is tangent to the circle at $B$.

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

b. Yes

c. Graph

d. 90$text{textdegree}$

b.
$$
moverset{frown}{BD}=mangle BCD=180text{textdegree}-angle A=180text{textdegree}-x
$$

$$
moverset{frown}{BZD}=360text{textdegree}-moverset{frown}{BD}=360text{textdegree}-(180text{textdegree}-x)=180text{textdegree}+x
$$

c.
$$
x=moverset{frown}{BZD}-180text{textdegree}=180text{textdegree}-moverset{frown}{BD}
$$

Thus $angle A$ and $moverset{frown}{BD}$ are supplementary, while $moverset{frown}{BZD}$ is 180$text{textdegree}$ increased by $angle A$.

d.
$$
x=moverset{frown}{BZD}-180text{textdegree}=180text{textdegree}-moverset{frown}{BD}
$$

b. $moverset{frown}{BZD}=180text{textdegree}+x$

c. Supplementary angles

d. $x=moverset{frown}{BZD}-180text{textdegree}=180text{textdegree}-moverset{frown}{BD}$

$$
x=90text{textdegree}+180text{textdegree}=270text{textdegree}
$$

$$
tan{24text{textdegree}}=dfrac{7}{y}
$$

Multiply both sides of the equation by $y$:

$$
ytan{24text{textdegree}}=7
$$

Divide both sides of the equation by $tan{48text{textdegree}}$:

$$
y=dfrac{7}{tan{24text{textdegree}}}approx 15.72
$$

We also need to determine the value of $x$, which is one of the angles in the quadrilateral with two right angles and an angle of $48text{textdegree}$.

Since a right angle is $90text{textdegree}$ and the sum of all angles in a quadrilateral needs to be equal to $360text{textdegree}$:

$$
begin{align*}
x&=360text{textdegree}-90text{textdegree}-90text{textdegree}-48text{textdegree}=132text{textdegree}
end{align*}
$$

$$
dfrac{x}{6}=dfrac{3}{x+2}
$$

Use cross multiplication:

$$
x^2+2x=18
$$

Subtract 18 from both sides of the equation:

$$
x^2+2x-18=0
$$

Determine the roots using the quadratic formula:

$$
x=dfrac{-2pm sqrt{2^2-4(1)(-18)}}{2(1)}=-1pm sqrt{19}
$$

Only a positive length makes sence:

$$
x=-1+sqrt{19}
$$

b. $y=15.72$, $x=132text{textdegree}$

c. $-1+sqrt{19}$

$$
(x-x_1)^2+(y-y_1)^2=r^2
$$

Thus the equation of this circle is then
$$
(x-4)^2+(y-2)^2=3^2
$$

$$
mangle QSO=dfrac{360text{textdegree}-180text{textdegree}-80text{textdegree}}{2}=dfrac{100text{textdegree}}{2}=50text{textdegree}
$$

$$
mangle QPO=mangle QSO=50text{textdegree}
$$

$$
mangle ONS=180text{textdegree}-63text{textdegree}-50text{textdegree}=67text{textdegree}
$$

$$
moverarc{PS}=2mangle SON=2(63text{textdegree})=126text{textdegree}
$$

$$
mangle PQN = 63text{textdegree}
$$

b. 50$text{textdegree}$

c. 67$text{textdegree}$

d. 126$text{textdegree}$

e. 54$text{textdegree}$

f. 63$text{textdegree}$

$$
AB=sqrt{5^2+12^2}=sqrt{169}=13
$$

$$
r=dfrac{13}{2}=6.5
$$

$$
mangle ABC=tan^{-1}dfrac{12}{5}approx 67text{textdegree}
$$

$$
moverarc{AC}=2(67text{textdegree})=134text{textdegree}
$$

b. 6.5

c. 67$text{textdegree}$

d. 134$text{textdegree}$

$$
(x-5)(x+2)=0
$$

Zero product property:

$$
x-5=0text{ or }x+2=0
$$

Solve each equation to $x$:

$$
x=5text{ or }x=-2
$$

$$
3x^2-4x+10=0
$$

Determine the roots using the quadratic formula:

$$
x=dfrac{4pm sqrt{(-4)^2-4(3)(10)}}{2(3)}=dfrac{2}{3}pm dfrac{sqrt{26}}{3}i
$$

$$
(x-8)(x-1)=0
$$

Zero product property:

$$
x-8=0text{ or }x-1=0
$$

Solve each equation to $x$:

$$
x=8text{ or }x=1
$$

$$
y^2-2y-15=0
$$

Factorize:

$$
(x-5)(x+3)=0
$$

Zero product property:

$$
x-5=0text{ or }x+3=0
$$

Solve each equation to $x$:

$$
x=5text{ or }x=-3
$$

b. $x=frac{2}{3}pm frac{sqrt{26}}{3}i$

c. $x=8$ or $x=1$

d. $x=5$ or $x=-3$

$$
dfrac{1}{4}x+2=12
$$

Subtract 2 from both sides of the equation:

$$
dfrac{1}{4}x=10
$$

Multiply both sides of the equation by 4:

$$
x=40
$$

Thus it will take 40 hours.

$$
h=5sin{72text{textdegree}}approx 4.76
$$

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

$$
dfrac{10cdot 4.76}{2}=23.8
$$

Since the decagon is made up out of 10 triangles, the area of the decagon is:

$$
10cdot 23.8=238units^2
$$

$$
mangle PSR=360text{textdegree}-140text{textdegree}=220text{textdegree}
$$

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

$$
mangle Q=360text{textdegree}-220text{textdegree}-40text{textdegree}-35text{textdegree}=65text{textdegree}
$$