$x=log (5^2)Rightarrow 10^x=25$

$y=2log (5)Rightarrow dfrac{y}{2}=log (5)Rightarrow 10^{y/2}=5Rightarrow 10^y=5^2=25$

$Rightarrow x=y$

$log (5^2)=2log (5)checkmark$

$log (7^3)stackrel{?}{=}3log (7)$

$x=log (7^3)Rightarrow 10^x=343$

$y=3log (7)Rightarrow dfrac{y}{3}=log (7)Rightarrow 10^{y/3}=7Rightarrow 10^y=7^3=343$

$Rightarrow x=y$

$log (7^3)=3log (7)checkmark$

$log (3^5)stackrel{?}{=}5log (3)$

$x=log (3^5)Rightarrow 10^x=243$

$y=5log (3)Rightarrow dfrac{y}{5}=log (3)Rightarrow 10^{y/5}=3Rightarrow 10^y=3^5=343$

$Rightarrow x=y$

$log (3^5)=5log (3)checkmark$

$log (2^8)stackrel{?}{=}8log (2)$

$x=log (2^8)Rightarrow 10^x=256$

$y=8log (2)Rightarrow dfrac{y}{8}=log (2)Rightarrow 10^{y/8}=2Rightarrow 10^y=2^8=256$

$Rightarrow x=y$

$log (2^8)=8log (2)checkmark$

$xlog (1.04)=log (2)$

$x=dfrac{log (2)}{log (1.04)}$

$$
xapprox 17.67
$$

$$
3^x=12
$$

$$
x^3=12
$$

The first equation is exponential. The second equation is cubic.

We solve the first equation. Take the $log$ of both sides.

$$
ln left(3^xright)=ln left(12right)
$$

Apply the rule $log _aleft(x^bright)=bcdot log _aleft(xright)$:

$$
xln left(3right)=ln left(12right)
$$

Isolate $x$:

$$
x=frac{ln left(12right)}{ln left(3right)}
$$

Use the calculator:

$$
x approx 2.26
$$

Now, let’s solve the second equation.

$$
x^3 = 12 Rightarrow x=sqrt[3]{12}
$$

Take the $log$ of both sides:

$$
ln left(1.04^xright) =ln left(2right)
$$

Apply the rule $log _aleft(x^bright)=bcdot log _aleft(xright)$:

$$
xln left(1.04right)=ln left(2right)
$$

$$
Rightarrow
$$

$$
x=frac{ln left(2right)}{ln left(1.04right)}
$$

$$
x approx 17.67
$$

b) We have $5^x=15$:
Take the $log$ of both sides:

$$
ln left(5^xright)=ln left(15right)
$$

Apply the rule $log _aleft(x^bright)=bcdot log _aleft(xright)$:

$$
xln left(5right)=ln left(15right)
$$

$$
Rightarrow
$$

$$
x=frac{ln left(15right)}{ln left(5right)}
$$

$$
x approx 1.68
$$

c) We have $3=8^x$:
Take the $log$ of both sides:

$$
ln left(3right)=ln left(8^xright)
$$

Apply the rule $log _aleft(x^bright)=bcdot log _aleft(xright)$:

$$
ln left(3right)=xln left(8right)
$$

$$
Rightarrow
$$

$$
x=frac{ln left(3right)}{3ln left(2right)}
$$

$$
xapprox0.528
$$

$-5-1<2x+1-1<5-1$

$-6<2x<4$

$-3<x<2$

$$
xin(-3,2)
$$

$3x-2leq -5$ or $3x-2geq 5$

$3x-2+2leq -5+2$ or $3x-2+2geq 5+2$

$3xleq -3$ or $3xgeq 7$

$xleq -1$ or $xgeq dfrac{7}{3}$

$$
xin(-infty,-1]cupleft[dfrac{7}{3},inftyright)
$$

b) $xin(-infty,-1]cupleft[dfrac{7}{3},inftyright)$

$2^x=30$

$$
2^4<2^x<2^5
$$

$xlog_{10} (2)=log_{10} (30)$

$x=dfrac{log_{10} (30)}{log_{10} (2)}$

$$
xapprox 4.91
$$

$5^x=200$

$log_{10} (5^x)=log_{10} (200)$

$xlog_{10} (5)=log_{10} (200)$

$x=dfrac{log_{10} (200)}{log_{10} (5)}$

$$
xapprox 3.29
$$

$b^x=y$

$log_{10} b^x=log_{10} y$

$xlog_{10} b=log_{10} y$

$$
x=dfrac{log_{10} y}{log_{10} b}
$$

b) $x=dfrac{log_{10} y}{log_{10} b}$

$$
dfrac{x^m}{x^n}=x^{m-n}
$$

$log (5cdot 6)=log (x)$

$log (30)=log (x)$

$$
x=30
$$

$log (5cdot 2)=log (x)$

$log (10)=log (x)$

$$
x=10
$$

$log (5cdot 5)=log (x)$

$log (25)=log (x)$

$$
x=25
$$

$log (10cdot 100)=log (x)$

$log (1000)=log (x)$

$$
x=1000
$$

$log (9cdot 11)=log (x)$

$log (99)=log (x)$

$$
x=99
$$

$log (mcdot n)=log (x)$

$log (mn)=log (x)$

$$
x=mn
$$

$log left(dfrac{20}{5}right)=log (x)$

$log (4)=log (x)$

$$
x=4
$$

$log left(dfrac{30}{3}right)=log (x)$

$log (10)=log (x)$

$$
x=10
$$

$log left(dfrac{5}{2}right)=log (x)$

$log (2.5)=log (x)$

$$
x=2.5
$$

$log left(dfrac{17}{9}right)=log (x)$

$$
x=dfrac{17}{9}
$$

$log left(dfrac{375}{17}right)=log (x)$

$$
x=dfrac{375}{17}
$$

$log left(dfrac{m}{n}right)=log (x)$

$$
x=dfrac{m}{n}
$$

Suppose you have some numbers $a, b, c$, where $a$ is neither 0 nor 1, and $a, b,$ and $c$ are strictly positive (not zero).

Then, for any values of $a, b,$ and $c$ in these conditions, it will be true that

$$
color[RGB]{150,0,0}log_a (b cdot c) = log_ab + log_ac
$$

$textbf{Example:}$

We wish to find the value of

$$
log_2{32}
$$

We know that $32 = 4 cdot 8$. So we can rewrite the expression as

$$
log_2(4 cdot 8)
$$

Now we have the logarithm of a product. We apply the property:

$$
log_2(4 cdot 8) = log_24 + log_28 = 2 + 3 = 5
$$

We can check that our result is correct, since $2^5 = 32$.

Suppose you have some numbers $a, b, c$, where $a$ is neither 0 nor 1, and $a, b,$ and $c$ are strictly positive (not zero).

Then, for any values of $a, b,$ and $c$ in these conditions, it will be true that

$$
color[RGB]{150,0,0}log_a left(frac{b}{c}right) = log_ab – log_ac
$$

$textbf{Example:}$

We wish to solve

$$
log_2{65536} – log_2{4096}
$$

Since we are subtracting two logarithms of the same base, we can turn the expression into the logarithm of a quotient.

$$
log_2{65536} – log_2{4096} = log_2 left(frac{65536}{4096}right)
$$

We can simplify the fraction and easily solve the logarithm:

$$
log_2 left(frac{65536}{4096}right) = log_216 = 4
$$

Suppose you have some numbers $a, b, c$, where $a$ and $b$ are neither 0 nor 1, and $a, b,$ and $c$ are strictly positive (not zero).

Then, for any values of $a, b,$ and $c$ in these conditions, it will be true that

$$
color[RGB]{150,0,0}log_ac = frac{log_bc}{log_ba}
$$

$textbf{Example:}$

We wish to solve the logarithm

$$
log_{27}243
$$

Note that both 27 and 243 are powers of 3. We can rewrite the logarithm in terms of base 3 using the Change of Base Property:

$$
log_{27}243 = frac{log_3{243}}{log_3{27}} = frac{5}{3}
$$

(e) $log left( frac{4 pi r^3}{3} right)$ (f) $log (6 cdot 10^{23})$

$$
n=b^y
$$

$log_b (m)=xlog_b (b)$

$log_b (m)=x$

$n=b^yRightarrow log_b (n)=log_b (b^y)$

$log_b (n)=ylog_b (b)$

$log_b (n)=y$

$mn=b^{x+y}Rightarrow log_b (mn)=log_b (b^{x+y})$

$log_b (mn)=(x+y)log_b (b)$

$log_b (mn)=x+y$

$y=log_b (n)$

$$
x+y=log_b (mn)
$$

$log_b left(dfrac{m}{n}right)=log_b (b^{x-y})$

$log_b left(dfrac{m}{n}right)=(x-y)log_b b$

$log_b left(dfrac{m}{n}right)=x-y$

$U=log_b (m)$.

$m^n=(b^U)^n$

$m^n=b^{Un}$

$log_b (m^n)=log_b (b^{Un})$

$log_b (m^n)=Unlog_b b$

$log_b (m^n)=Un$

$$
log_b (m^n)=nlog_b (m)
$$

$3log_5 (2)=x$

$$
dfrac{3log_{10} (2)}{log_{10} (5)}=x
$$

$$
=dfrac{5log_2 (2)}{2log_2 (2)}=dfrac{5}{2}
$$

$$
=dfrac{5log (2)}{2log (2)}=dfrac{5}{2}
$$

We use the change of base formula:

The vertex is $(2,3)$          (Given)

The parabola passes the point $(0, 0)$          Given

Substituting in the general form for the vertex and the given point.

$0=a(0-2)^2+3$

$4a=-3$

$a=-dfrac {3}{4}$

The equation of the parabola is:

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

$$
x_1=2; y_2=392
$$

$a=200$

$200b^2=392$

$b^2=dfrac{392}{200}$

$b^2=dfrac{49}{25}$

$$
b=dfrac{7}{5}=1.4
$$

$(1.4)^x=dfrac{5}{200}$

$(1.4)^x=0.25$

$log (1.4)^x=log 0.025$

$xlog 1.4=log 0.025$

$x=dfrac{log 0.025}{log 1.4}$

$$
xapprox-11
$$

Until April 1st: $31-10=21$ days

Until May 1st: $21+30=51$ days

Until June 1st: $51+31=82$ days

Until July 1st: $82+30=112$ days

Until August 1st: $112+31=143$ days

Until September 1st: $143+31=174$ days

Until October 1st: $174+30=204$ days

Until November 1st: $204+31=235$ days

Until December 1st: $235+30=265$ days

$$
(23,392)
$$

$b^2=1.96$

$b=sqrt{1.96}$

$$
b=1.4
$$

$a=dfrac{200}{(1.4)^{21}}$

$$
aapprox 0.1707
$$

$$
(4,11.25)
$$

$$
begin{cases}
ab^3=12.5-10\
ab^4=11.25-10
end{cases}
$$

$$
begin{cases}
ab^3=2.5\
ab^4=1.25
end{cases}
$$

$dfrac{ab^4}{ab^3}=dfrac{1.25}{2.5}$

$b=0.5$

$a(0.5)^3=2.5$

$a=dfrac{2.5}{0.125}$

$a=20$

$$
y=20(0.5)^x+10
$$

$A=40,000$

$n=1$

$$
t=8
$$

$$
A=P(1+r)^t
$$

$dfrac{40,000}{1000}=(1+r_1)^8$

$40=(1+r_1)^8$

$1+r_1=sqrt[8]{40}$

$r_1=1.5858-1$

$r_1=0.5868$

$$
r_1=58.58%
$$

$A=18,400$

$n=1$

$$
t=20
$$

$dfrac{18,400}{7800}=(1+r_2)^{20}$

$2.359=(1+r_2)^{20}$

$1+r_2=sqrt[20]{2.359}$

$r_2=1.0438-1$

$r_2=0.0438$

$$
r_2=4.38%
$$

$$
(3,135)
$$

$$
begin{cases}
ab^1=60\
ab^3=135
end{cases}
$$

$dfrac{ab^3}{ab^1}=dfrac{135}{60}$

$b^2=dfrac{9}{4}$

$b=dfrac{3}{2}$

$$
a=dfrac{60}{dfrac{3}{2}}=60cdot dfrac{2}{3}=40
$$

$(1.5)^x=dfrac{1}{40}$

$log_{10} (1.5)^x=log_{10} 0.025$

$xlog_{10} (1.5)=log_{10} 0.025$

$x=dfrac{log_{10} 0.025}{log_{10} (1.5)}$

$$
xapprox -9
$$

$t=20$

$t_1=6$

$$
A_1=132
$$

$$
A=P(1+r)^t
$$

$132=100(1+r)^6$

$dfrac{132}{100}=(1+r)^6$

$1.32=(1+r)^6$

$1+r=sqrt[6]{1.32}$

$r=1.0474-1$

$$
r=0.0474=4.74%
$$

$A=100(1+0.0474)^{20}$

$A=100(1.0474)^{20}$

$$
Aapprox 252.49
$$

$$
Aapprox $252.49
$$

where

$T$=the temperature of the cup at time $t$

$t$=time

$T_s$=the temperature of the room

$T_0$=the initial temperature of the cocoa

$k$=a cooling constant, specific to the cocoa

$T_0=373.15$

$T(t)=293.15+(373.15-293.15)e^{-0.0015t}$

$$
T(t)=293.15+80e^{-0.0015t}
$$

Sketch the graph in Desmos.

We graph both functions and shade the region above the first function and below the second and intersect them to get the solution region:

$$
AB = sqrt{(5-5)^2+(2-(-8))^2}
$$

$$
AB =sqrt{0+100}
$$

$$
AB= 10
$$

We know:

$$
P_{triangle ABC} = dfrac{AB cdot h}{2}
$$

From the graph we see $h=5$.
Substitute the values and we have:

$$
P_{triangle ABC} = dfrac{10 cdot 5 }{2}
$$

$$
P_{triangle ABC} = colorbox{yellow}{25}
$$

$$
i^3=icdot i^2=icdot (-1)=-i
$$

$$
i^4=i^2cdot i^2=(-1)cdot (-1)=1
$$

a) $(x+4)(2x-5)=0qquadqquadqquadqquadqquad$ $color{#c34632} text{[apply the zero product property]}$

Set each factor equal to zero and solve:

$bullet,,$ $x+4=0Rightarrow color{#4257b2} text{$x=-4$}$

$bullet,,$ $2x-5=0Rightarrow 2x=5Rightarrow color{#4257b2} text{$x=dfrac{5}{2}$}$

The solutions of the initial equation are: $color{#4257b2} text{$,,,x=-4$}$ and $color{#4257b2} text{$x=dfrac{5}{2}$}$

Set each factor equal to zero and solve:

$bullet,,$ $x+4=0Rightarrow color{#4257b2} text{$x=-4$}$

$bullet,,$ $x^2-5x+6=0qquadqquadqquadqquadqquad$ $color{#c34632} text{[$a=1$ , $b=-5$ , $c=6$]}$

$color{#c34632} text{Apply the quadratic formula:}$

$color{#c34632} text{$x_{1,2}=dfrac{-bpm sqrt{b^2-4ac}}{2a}$}$

$Rightarrow x_{1,2}=dfrac{5pm sqrt{(-5)^2-4(1)(6)}}{2(1)}$

$Rightarrow x_{1,2}=dfrac{5pm sqrt{25-24}}{2}$

$Rightarrow x_{1,2}=dfrac{5pm 1}{2}$

$Rightarrow color{#4257b2} text{$x=2,,$}$ or $color{#4257b2} text{$,,x=3$}$

The solutions of the initial equation are: $color{#4257b2} text{$,,,x=-4$}$ , $color{#4257b2} text{$x=2$}$ and $color{#4257b2} text{$x=3$}$

$color{#c34632} text{[apply the zero product property]}$

Set each factor equal to zero and solve the resulting equations as shown below:

$bullet,,$ $3x=0Rightarrow color{#4257b2} text{$x=0$}$

$bullet,,$ $x+1=0Rightarrow color{#4257b2} text{$x=-1$}$

$bullet,,$ $2x-7=0Rightarrow 2x=7Rightarrow color{#4257b2} text{$x=dfrac{7}{2}$}$

$bullet,,$ $3x+4=0Rightarrow 3x=-4Rightarrow color{#4257b2} text{$x=-dfrac{4}{3}quadquad$}$ $color{#4257b2} text{Multiplicity $2$}$

$bullet,,$ $x-13=0Rightarrow color{#4257b2} text{$x=13$}$

$bullet,,$ $x+7=0Rightarrow color{#4257b2} text{$x=-7$}$

The solutions of the initial equation are:

$color{#4257b2} text{$x=0$}$ , $color{#4257b2} text{$x=-1$}$ , $color{#4257b2} text{$x=dfrac{5}{2}$}$ , $color{#4257b2} text{$x=dfrac{7}{2}$ , $x=-dfrac{4}{3}$ , $x=13,,$}$ and $color{#4257b2} text{$,,x=-7$}$

$$
y=ab^x+17
$$

$$
begin{cases}
dfrac{a}{b}=27-17=10\
a=24-17=7
end{cases}
$$

$dfrac{a}{b}=10$

$dfrac{7}{b}=10$

$$
b=dfrac{7}{10}=0.7
$$

$7left(0.7right)^x+17=37$

$7left(0.7right)^x=37-17$

$7left(0.7right)^x=20$

$(0.7)^x=dfrac{20}{7}$

$log (0.7)^x=log dfrac{20}{7}$

$xlog (0.7)=log dfrac{20}{7}$

$x=dfrac{log dfrac{20}{7}}{log (0.7)}$

$$
xapprox -2.94
$$

$$
6h12′-2h56.4’=5h72′-2h56.4’=3h 15.4′
$$

$$
A(t)=P(0.8)^t
$$

$P$=the initial value of the car.

After the first year its value will be $100%-20%=80%$ from the initial value, after the second year its value will be $80%$ from the value after the first year and so on. We can write the model:

$(0.8)^t=dfrac{6000}{23,500}$

$(0.8)^t=0.255319$

$log_{10} (0.8)^t=log_{10} 0.255319$

$tlog_{10} (0.8)=log_{10} 0.255319$

$t=dfrac{log_{10} 0.255319}{log_{10} (0.8)}$

$tapprox 6$ years

$P_0=dfrac{23,500}{(0.8)^{2.7}}$

$$
P_0=42,926.44
$$

$$
x=pm 2.236
$$

$x^{19}=796,262,400,000$

$x=sqrt[19]{796,262,400,000}$

$$
xapprox 4.230
$$

$1=100x^4$

$x^4=dfrac{1}{100}$

$x=pmdfrac{1}{sqrt[4]{100}}$

$$
x=pm 0.316
$$

$x=4.021-2$

$$
x=2.021
$$

$16(x-2)^{25}=6,180,196$

$(x-2)^{25}=dfrac{6,180,196}{16}$

$(x-2)^{25}=386,262.25$

$x-2=sqrt[25]{386,262.25}$

$x=2+1.673$

$$
x=3.673
$$

b. The vertical equation has equation (about) $x=2$.

c. Since $ln{1}=0$, a possible function is $y=ln{(x-2)}$ (thus the parent function translates to the right by 2 units).

b. $x=2$

c. $y=ln{(x-2)}$

$f$(x) = 3(x – 4)$^{2}$ – 5

g(x) = 2x$^{2}$ – 3x – 5

2) $2$ sides of a triangle and $1$ angle

3) $1$ side and $2$ angles of a triangle

4) all the $3$ angles of a triangle

2) $2$ sides of a triangle and $1$ angle
…

$AB^2+4^2=5^2$

$AB^2=25-16$

$AB^2=9$

$$
AB=3
$$

$AC=18sin 30text{textdegree}$

$AC=18cdot dfrac{1}{2}$

$AC=9$

$AB^2+AC^2=BC^2$

$AB^2=18^2-9^2$

$AB^2=243$

$AB=sqrt{243}$

$$
AB=9sqrt 3
$$

$a^2=14^2+20^2-2(14)(20)cdot dfrac{1}{2}$

$a^2=316$

$a=sqrt{316}$

$aapprox 17.78$

$14<17.78<20$

$14^2+17.78^2=512not=20^2Rightarrow$the triangle is not right

$5sin B=7sin 40text{textdegree}$

$sin B=dfrac{7cdot 0.64278761}{5}$

$sin B=0.8999$

$Bapprox 64.1text{textdegree}$

$C=180text{textdegree}-40text{textdegree}-64.1text{textdegree}=75.9text{textdegree}Rightarrow$the triangle is not right

$sin 30text{textdegree}=dfrac{BD}{14}$

$dfrac{1}{2}cdot 14=BD$

$$
BD=7
$$

$AD^2+7^2=14^2$

$AD^2=196-49$

$AD^2=147$

$$
AD=7sqrt 3
$$

$CD^2+AD^2=AC^2$

$13^2+147=AC^2$

$AC^2=316$

$AC=sqrt{316}$

$$
ACapprox 17.78
$$

$sin C=dfrac{7sqrt 3}{17.78}$

$sin Capprox 0.6819$

$$
Capprox 43text{textdegree}
$$

$=180text{textdegree}-60text{textdegree}-43text{textdegree}$

$$
=77text{textdegree}
$$

$AD=dfrac{1}{2}cdot 18$

$$
AD=9
$$

$CD^2=18^2-9^2$

$CD^2=275$

$CD=sqrt{243}$

$$
CD=9sqrt 3
$$

$BD^2+AD^2=AB^2$

$9^2+9^2=AB^2$

$162=AB^2$

$$
AB=9sqrt 2
$$

$$
=180text{textdegree}-45text{textdegree}-30text{textdegree}=105text{textdegree}
$$

$AD=sin 48text{textdegree}cdot 5$

$AD=0.74314483cdot 5$

$$
ADapprox 3.72
$$

$ACB=180text{textdegree}-48text{textdegree}-29text{textdegree}$

$$
=103text{textdegree}
$$

$AC=dfrac{3.72}{sin 77text{textdegree}}=dfrac{3.72}{0.97437006}$

$$
ACapprox 3.82
$$

$sin CAD=dfrac{CD}{AC}$

$CD=sin 13text{textdegree}cdot 3.82=0.22495105cdot 3.82$

$$
CDapprox 0.86
$$

$BD=cos 48text{textdegree}cdot 5=0.66913061cdot 5$

$$
BDapprox 3.35
$$

$$
BC=2.49
$$

$$
=170text{textdegree}
$$

$$
log _{10}left(8right)+log _{10}left(125right)=log _{10}left(8cdot :125right)
$$

Multiply.

$$
log _{10}left(8cdot :125right)=log _{10}left(1000right)
$$

Rewrite.

$$
log _{10}left(1000right)=log _{10}left(10^3right)
$$

Apply the rule $log _aleft(x^bright)=bcdot log _aleft(xright)$:

$$
log _{10}left(10^3right)=3log _{10}left(10right)
$$

Apply the rule $log _aleft(aright)=1$:

$$
3log _{10}left(10right)=3cdot :1 = 3
$$

$$
log _{25}left(125right)=log _{5^2}left(125right)
$$

Apply the rule $log _{a^b}left(xright)=frac{1}{b}log _aleft(xright)$:

$$
log _{5^2}left(125right)=frac{1}{2}log _5left(125right)
$$

Rewrite.

$$
frac{1}{2}log _5left(125right)=frac{1}{2}log _5left(5^3right)
$$

Apply the rule $log _aleft(a^xright)=x$:

$$
frac{1}{2}log _5left(5^3right)=frac{1}{2}cdot :3=frac{3}{2}
$$

$$
log _{10}left(5right)+log _{10}left(20right) =log _{10}left(5cdot :20right)
$$

Multiply.

$$
log _{10}left(5cdot :20right)=log _{10}left(100right)
$$

Apply the rule $log _aleft(x^bright)=bcdot log _aleft(xright)$:

$$
log _{10}left(100right)=log _{10}left(10^2right)=2log _{10}left(10right)
$$

Apply the rule $log _aleft(aright)=1$:

$$
2log _{10}left(10right)=2cdot :1 = 2
$$

Apply the rule $a^{log _aleft(bright)}=b$:

$$
7^{log _7left(12right)}=12
$$

Mean score of French exam ($bar x_f$) = 81
Standard Deviation ($sigma_f$) = 5
Mean score of Spanish exam ($bar x_s$) = 72
Standard Deviation ($sigma_s$) = 12

In order to earn honors in language at graduation, a student must score in the 90th percentile on all their language finals.

Brynne has scored 88 in both French and Spanish finals. We have to see whether he has earned his honors or not.

It is pertinent to note,
$$z=dfrac{x-bar{x}}{sigma}$$

Now we will use the normal distribution table,

We get
$z<1.4 = 0.91924$

This implies Brynne falls in the $91.92^{text{th}}$ which is greater than the $90^{text{th}}$ percentile.Therefore, Brynne has qualified for honors in French.

Now we will use the normal distribution table,

We get
$z<1.333 = 0.90879$

This implies Brynne falls in the $90.88^{text{th}}$ which is greater than the $90^{text{th}}$ percentile.Therefore, Brynne has qualified for honors in Spanish as well.

Since, Brynne's score for both the exams is 88 which is above the $90^{text{th}}$ percentile, he has earned the language honor at graduation.

$$
a)
$$

$color{#c34632} text{$y=7+2x^2+4x-5$}$

First write this quadratic equation in the standard form $color{#c34632} text{$,,y=ax^2+bx+c,,$}$ as shown:

$y=7+2x^2+4x-5$

$Rightarrow color{#4257b2} text{$y=2x^2+4x+2$}$

$y=0$

$Rightarrow 2x^2+4x+2=0qquadqquadqquadqquadqquad$ $color{#c34632} text{[factor out $2$]}$

$Rightarrow 2(x^2+2x+1)=0qquadqquadqquadqquad$ $color{#c34632} text{[factor the perfect square trinomial]}$

$Rightarrow 2(x+1)^2=0$

$Rightarrow (x+1)^2=0$

$Rightarrow x+1=0qquadqquadqquadqquadqquadqquad$ $color{#c34632} text{[subtract $1$ in both sides]}$

$Rightarrow color{#4257b2} text{$x=-1$}$

The $x$-intercept of the graph of this equation is the point $color{#4257b2} text{$,,(-1 , 0)$}$.

$bullet,,$For the $y$-intercept set $color{#c34632} text{$,,x=0,,$}$ and solve the resulting equation for $color{#c34632} text{$,,”y”$}$:

$y=2x^2+4x+2qquadqquadqquadqquadqquadqquad$ $color{#c34632} text{[set $x=0$]}$

$Rightarrow y=2(0)^2+4(0)+2qquadqquadqquadqquadqquad$

$Rightarrow color{#4257b2} text{$y=2qquadqquadqquadqquad$}$

The $y$-intercept of the graph of this equation is the point $color{#4257b2} text{$,,(0 , 2)$}$.

$color{#4257b2} text{$(h,k)=bigg(dfrac{-b}{2a} ,,, fbigg(dfrac{-b}{2a}bigg)bigg)$}$

In this case we have:

$color{#4257b2} text{$f(x)=2x^2+4x+2$}qquadqquadqquadqquadqquad$ $color{#c34632} text{[$a=2$ , $b=4$ , $c=2$]}$

$h=dfrac{-b}{2a}=dfrac{-4}{2(2)}=dfrac{-4}{4}=color{#4257b2} text{$-1$}$

$k=fbigg(dfrac{-b}{2a}bigg)=f(-1)=2(-1)^2+4(-1)+2=color{#4257b2} text{$0$}$

$bullet,,$The vertex of the graph of this function is the point $color{#4257b2} text{$,,(h,k)=(-1,0)$}$.

$bullet,,$Write this equation in the vertex form $color{#c34632} text{$,,y=a(x-h)^2+k,,$}$ as shown below:

$y=2x^2+4x+2qquadqquadqquadqquadqquadqquad$ $color{#c34632} text{[factor out $2$]}$

$Rightarrow y=2(x^2+2x+1)qquadqquadquadquadquad$ $color{#c34632} text{[factor the perfect square trinomial]}$

$$
Rightarrow color{#4257b2} text{$y=2(x+1)^2+0$}
$$

$color{#c34632} text{$x^2=2x+x(2x-4)+y$}$

First write this quadratic equation in the standard form $color{#c34632} text{$,,y=ax^2+bx+c,,$}$ as shown:

$x^2=2x+x(2x-4)+yqquadqquadqquadqquadqquad$ $color{#c34632} text{[solve for $y$]}$

$Rightarrow y=x^2-2x-x(2x-4)$

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

$Rightarrow color{#4257b2} text{$y=-x^2+2x$}$

$y=0$

$Rightarrow -x^2+2x=0qquadqquadqquadqquadqquad$ $color{#c34632} text{[factor out $x$]}$

$Rightarrow x(2-x)=0qquadqquadqquadqquad$ $color{#c34632} text{[zero product property]}$

$Rightarrow x=0quad$ or $quad 2-x=0$

$Rightarrow color{#4257b2} text{$x=0quad$}$ or $color{#4257b2} text{$quad x=2$}$

The $x$-intercepts of the graph of this equation are the points $color{#4257b2} text{$,,(0 , 0),,$}$ and $color{#4257b2} text{$,,(2,0),,$}$.

$bullet,,$For the $y$-intercept set $color{#c34632} text{$,,x=0,,$}$ and solve the resulting equation for $color{#c34632} text{$,,”y”$}$:

$y=-x^2+2xqquadqquadqquadqquadqquadqquad$ $color{#c34632} text{[set $x=0$]}$

$Rightarrow y=-(0)^2+2(0)qquadqquadqquadqquadqquad$

$Rightarrow color{#4257b2} text{$y=0qquadqquadqquadqquad$}$

The $y$-intercept of the graph of this equation is the point $color{#4257b2} text{$,,(0 , 0)$}$.

$color{#4257b2} text{$(h,k)=bigg(dfrac{-b}{2a} ,,, fbigg(dfrac{-b}{2a}bigg)bigg)$}$

In this case we have:

$color{#4257b2} text{$f(x)=-x^2+2x$}qquadqquadqquadqquadqquad$ $color{#c34632} text{[$a=-1$ , $b=2$ , $c=0$]}$

$h=dfrac{-b}{2a}=dfrac{-2}{2(-1)}=dfrac{-2}{-2}=color{#4257b2} text{$1$}$

$k=fbigg(dfrac{-b}{2a}bigg)=f(1)=-1^2+2(1)=color{#4257b2} text{$1$}$

$bullet,,$The vertex of the graph of this function is the point $color{#4257b2} text{$,,(h,k)=(1,1)$}$.

$bullet,,$Write this equation in the vertex form $color{#c34632} text{$,,y=a(x-h)^2+k,,$}$ as shown below:

$y=-x^2+2xqquadqquadqquadqquadqquadqquad$ $color{#c34632} text{[factor out $-1$]}$

$Rightarrow y=-(x^2-2x)qquadqquadquadquadquad$ $color{#c34632} text{[add and subtract $1$ inside the parenthesses]}$

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

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

$color{#c34632} text{[factor the perfect square trinomial]}$

$$
Rightarrow color{#4257b2} text{$y=-(x-1)^2+1$}
$$

The total number of goats is 6. So,

Goats covered the frolic area is
$$begin{aligned}
6times1.1 &= 6.6\
end{aligned}$$
Since the area is 74.23 square meters.

Thus, yes this pen design meets animal rights advocate criteria.

$B<C<A$

$$
AC<AB<BC
$$

$AD=sin 42text{textdegree}cdot 4=0.66913061cdot 4$

$$
ADapprox 2.68
$$

$sin 26text{textdegree}=dfrac{2.68}{x}$

$x=dfrac{2.68}{0.43837115}$

$$
xapprox 6.11
$$

$CD=cos 42text{textdegree}cdot 4=0.74314483cdot 4$

$CDapprox 2.97$

$cos B=dfrac{BD}{AB}$

$BD=cos 26text{textdegree}cdot 6.11=0.89879405cdot 6.11$

$BDapprox 5.49$

$$
approx 11.34
$$

b) $x=6.11$

c) $11.34$

$$
sin 24text{textdegree}=dfrac{h}{25}
$$

$$
h=12sin 58text{textdegree}=12cdot 0.8480481approx 10.177
$$

$$
h=25sin 24text{textdegree}=25cdot 0.40673664approx 10.169
$$

$$
sin B=dfrac{h}{a}
$$

$$
h=asin B
$$

$$
dfrac{sin B}{b}=dfrac{sin A}{a}
$$

$$
dfrac{sin B}{b}=dfrac{sin A}{a}=dfrac{sin C}{c}
$$

$xsin A=BCsin B$

$$
x=dfrac{BCsin B}{sin A}=dfrac{10cdot 0.8660254}{0.46947156}=18.45
$$

Based on the image, we solve problems.

We know that the area of triangle is:

$$
P = dfrac{b cdot h_b}{2}
$$

So, in this case:

$$
P= dfrac{bcdot h}{2}
$$

$$
sin C = dfrac{h}{a}
$$

$$
h = a sin C
$$

$$
P = dfrac{ba sin C}{2}
$$

We compute tha angle between $x$ and the side of $10$:

$$
180 text{textdegree} – 28 text{textdegree} -120 text{textdegree} = 32 text{textdegree}
$$

We determine the triangle’s area using the sinus formula in $c$:

$$
P = dfrac{18.45 cdot 10 cdot sin 32 text{textdegree} }{2} = dfrac{18.45 cdot 10 cdot 0.53}{2}
$$

Therefore,

$$
P approx 48.89
$$

$Pcong RRightarrow sin P=sin R$

$$
dfrac{sin P}{QR}=dfrac{sin R}{QP}Rightarrow QR=QP
$$

$PRsin R=QPsin Q$

$PR=dfrac{QPsin Q}{sin R}$

$$
=dfrac{5cdot sin 84text{textdegree}}{sin 48text{textdegree}}=dfrac{5cdot 0.9945219}{0.74314483}approx 6.69
$$

b) $PR=6.69$ mm

$B(4,2499)$

$$
y=-1
$$

$ab^{4}+k=2499$

$ab^{4}-1=2499$

$ab^{-1}=-0.2+1$

$ab^{4}=2499+1$

$ab^{-1}=0.8$

$ab^{4}=2500$

$b^5=3125$

$b=sqrt[5]{3125}$

$$
b=5
$$

$a=dfrac{2500}{625}$

$$
a=4
$$

$B(4,7.1536)$

$$
y=7
$$

$ab^4+7=7.1536$

$ab^2=7.96-7$

$ab^4=7.1536-7$

$ab^2=0.96$

$ab^4=0.1536$

$dfrac{ab^4}{ab^2}=dfrac{0.1536}{0.96}$

$b^2=0.16$

$b=sqrt{0.16}$

$b=0.4$

$a=dfrac{0.96}{0.16}$

$$
a=6
$$

b) $y=6(0.4)^x+7$

$log_x (5)=b$

$log_x (7)=c$

$=log_x (2)+2log_x (5)$

$$
=a+2b
$$

$=3log_x (2)+log_x (7)$

$$
=3a+c
$$

b) $2c$

c) $a+2b$

d) $3a+c$

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

$$
y=-(x-1)^2+20
$$

$$
y=-3|x+71|
$$

$left(dfrac {1}{2} right)^{3(2x-3)}=left(dfrac {1}{2} right)^{(x+2)}$          (Unifying base)

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

$6x-9=x+2$

$6x-x=9+2$          (Grouping similar terms)

$5x=11$

$$
x=dfrac {11}{5}
$$

Write an equation relating $c,y$ and $DA$:
$$c^2=y^2+(b-x)^2$$
Then, we isolate $y^2$ in both equations:
$$begin{aligned}
y^2&=a^2-x^2\
y^2&=c^2-(b-x)^2
end{aligned}$$
Expand the second equation:
$$begin{aligned}
a^2-x^2&=c^2-(b-x)^2\
a^2-x^2&=c^2-b^2+2bx-x^2
end{aligned}$$
Therefore, it follows that:
$$a^2+b^2-c^2=2bx$$

$$cos C=dfrac{x}{a} implies x=acos C$$
We replace $x$ in the equation:
$$a^2+b^2-c^2=2bacos C$$
Solve for $c^2$ and we get the Law of Cosines.
$$c^2=a^2+b^2-2abcos C$$

$$
x=tan^{-1}{dfrac{5}{9}}
$$

We can use the $textbf{Law of cosines. }$

$$
c^2 = a^2+b^2 – 2abcos c
$$

$$
begin{align*}
10^2 &= 7^2 + 8^2 – 2cdot 7 cdot 8 cos alpha \
100 &= 49 + 64 – 112 cos alpha && text{Calculate} \
100 &= 113 – 112 cos alpha \
112 cos alpha &= 113 – 110 \
cos alpha &= dfrac{3}{112} &&text{Isolate $cos alpha$} \
cos alpha &approx 0.02679 \
alpha &approx 88.5text{textdegree}
end{align*}
$$

$$
begin{align*}
7^2 &= 10^2 + 8^2 – 2cdot 10 cdot 8 cos B \
49 &= 100+64 – 160 cos B && text{Calculate.} \
160 cos B &= 164 – 49 \
160 cos B&= 115 \
cos B &= dfrac{115}{160} &&text{We need to isolate $cos B$} \
cos B &= 0.7188 \
B &approx 44 text{textdegree}
end{align*}
$$

$$
A = 180 text{textdegree} – 88.5 text{textdegree} -44 text{textdegree} = 47.5 text{textdegree}
$$

b) $47.5$

The Law of Cosines gives a rule by which we can compute one side of a triangle when we are given the opposite angle and the sides near it.

$AC^2=7^2+5^2-2(7)(5)cos 60text{textdegree}$

$AC^2=49+25-70cdotdfrac{1}{2}$

$AC^2=39$

$$
AC=sqrt{39}approx 6.24
$$

$A=180text{textdegree}-B-C$

$A=180text{textdegree}-33text{textdegree}-64text{textdegree}$

$$
A=83text{textdegree}
$$

$xsin B=ACsin A$

$x=dfrac{ACsin A}{sin B}$

$x=dfrac{3.5cdot sin 83text{textdegree}}{sin 33text{textdegree}}$

$x=dfrac{3.5cdot 0.99254615}{0.54463904}$

$$
xapprox 6.38
$$

$$
=dfrac{6.38cdot 3.5cdot 0.89879405}{2}approx 10.04
$$

b) $10.04$

$$
approx 36.73
$$

b) $y=36.73$

$=dfrac{165cdot 100cdot sin 43text{textdegree}}{2}=dfrac{165cdot 100cdot 0.68199836}{2}$

$approx 5626.49$ square miles

$$
f(50)=25.25
$$

$$
begin{cases}
ab=1.3\
ab^{50}=25.05
end{cases}
$$

$b^{49}=19.269231$

$B=sqrt[49]{19.269231}$

$bapprox 1.06$

$a=dfrac{1.3}{1.06}$

$$
aapprox 1.23
$$

b) $$747.60$

$$
g(x)=sqrt[3]{x+2}
$$

$sqrt[3] x=sqrt[3]{x+2}$

$(sqrt[3] x)^3=(sqrt[3]{x+2})^2$

$x=x+2$

$$
0=2
$$

$x=y^2+3$

$y^2=x-3$

$y=sqrt{x-3}$

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

We determine its inverse:

$x=left(dfrac{1}{4}y+6right)^3$

$sqrt[3] x=dfrac{1}{4}y+6$

$4sqrt[3] x=y+24$

$$
y=4sqrt[3] x-24
$$

$x=sqrt{5y-6}$

$x^2=(sqrt{5y-6})^2$

$x^2=5y-6$

$5y=x^2+6$

$y=dfrac{x^2+6}{5}$

$$
f^{-1}(x)=dfrac{x^2+6}{5}
$$

We determine its inverse:

b) $f^{-1}(x)=4sqrt[3] x-24$

c) $f^{-1}(x)=dfrac{x^2+6}{5}$

$5sin x=7sin 40text{textdegree}$

$sin x=dfrac{7sin 40text{textdegree}}{5}$

$sin x=dfrac{7cdot 0.64278761}{5}$

$sin xapprox 0.8999$

$sin^{-1} 0.8999approx 64.1text{textdegree}$

$$
x=64.1text{textdegree}
$$

$$begin{aligned}
A &= 180degree − 30degree − 45degree\\
A &= 105°\\
end{aligned}$$


$$begin{aligned}
180degree &= 60degree + 75degree + 45degree\\
180° &= 180degree \\
end{aligned}$$
The solution are all the similar triangles.

$x^2=356-320cdot 0.74314483$

$x^2=118.19365$

$xapprox sqrt{118.19365}$

$$
xapprox 10.87
$$

$x=9sin 56text{textdegree}$

$x=9cdot 0.82903757$

$$
xapprox 7.46
$$

$dfrac{sin 9text{textdegree}}{x}=dfrac{sin 120text{textdegree}}{25}$

$xsin 120text{textdegree}=25sin 9text{textdegree}$

$x=dfrac{25cdot 0.15643447}{0.8660254}$

$$
xapprox 4.52
$$

b) $7.46$

c) $4.52$

$17sin x=15sin 41text{textdegree}$

$sin x=dfrac{15cdot 0.65605903}{17}$

$sin xapprox 0.57887561$

$$
sin^{-1} 0.57887561approx 35.4text{textdegree}
$$

$x_2=180text{textdegree}-35.4text{textdegree}=144.6text{textdegree}$

$B+C=x+41text{textdegree}$

$x_1=35.4text{textdegree}$

$B+C=35.4text{textdegree}+41text{textdegree}=76.4text{textdegree}180text{textdegree}$

We can use the law of Cosine to find the length of the BC side fence that needs to be replaced.

$$begin{aligned}
BC^{2} &= AB^{2} + AC^{2} – 2 cdot AB cdot ACcdot cos A\\
BC^{2} &= (116)^{2} + (224)^{2} – 2 times 116 times224 times Cos(58degree)\\
BC^{2} &= 13456 + 50176 – 51968 times 0.5299\\
BC^{2}&= 63632 – 26589.229\\
BC^{2} &= 36093.156\\
end{aligned}$$

Squaring both sides, we get,

$$begin{aligned}
BC &= sqrt {36093.156}\\
BC &approx 189.98 text{ feet}
end{aligned}$$

$$begin{aligned}
Area &= dfrac {1}{2} times AB times AC times sin A\\
&= dfrac {116 times 224 times sin (58degree)}{2}\\
&= dfrac {116 times 224 times 0.848048}{2}\\
Area &approx 11017.84\
end{aligned}$$

$xleq -4$ or $xgeq 2$

$x+3=0Rightarrow x_2=-3$

$f(0)=0^2-9=-00$

$$
(-3,3)
$$

From the above calculation we can see that the percentile is increasing. Thus, we can say that the advice by Duncan is right.

$x+1=243$

$x=243-1$

$$
x=242
$$

$$
x=4
$$

$a^{log_a (x)}=x$.

b) $x=242$

c) $x=4$

d) $x=7$

From the graph we see:
The parabola is shifted $3$ units to the left.

$$
sqrt{5-2x}+7=4
$$

Subtract $7$ from both sides

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

Squaring both sides

$$
left(sqrt{5-2x}right)^2=left(-3right)^2
$$

$$
5-2x=9
$$

$$
-2x=4
$$

$$
x=-2
$$

Replace $x$ with $-2$ in the original equation

$$
sqrt{5+4}+7=4
$$

$$
10=4
$$

which is a contradiction, then the equation has no real solution

$ACsin C=ABsin B$

$AC=dfrac{ABsin B}{sin C}=dfrac{684cdot sin 79text{textdegree}}{sin 53text{textdegree}}$

$=dfrac{684cdot 0.98162718}{0.79863551}$

$$
approx 840.73
$$

$=800^2+694^2-2cdot 800cdot 694cdot cos 68text{textdegree}$

$=1,121,636-1,110,400cdot 0.37460659$

$=705,672.84$

$$
AC=sqrt{705672.84}approx840.04
$$

Yes

$BCsin B=ACsin A$

$BC=dfrac{ACsin A}{sin B}=dfrac{2cdotsin 97text{textdegree}}{sin 23text{textdegree}}$

$=dfrac{2cdot 0.99254615}{0.39073113}$

$$
approx 5.08
$$

$=97^2+138^2-2cdot 97cdot 138cdot cos 28text{textdegree}$

$=28,453-26,772cdot 0.88294759$

$=4814.7271$

$$
PE=sqrt{4814.7271}approx 69.4
$$

$$
BC=sqrt{5467.54}approx 73.9
$$

First we determine $BC$ using the Pythagorean Theorem:

$=7099.7$

$$
AB=sqrt{7099.7}approx 84.3
$$

$AB=FBtan 38text{textdegree}=253cdot 0.78128563$

$$
=197.67
$$

First we determine $AB$ using the tangent function:

$BC=FBtan 46text{textdegree}=253cdot 1.03553031$

$$
=261.99
$$

$$begin{aligned}
BC^{2} &= AB^{2} + AC^{2} – 2cdot ABcdot ACcdot cos A\\
BC^{2} &= (5)^{2} + (7.4)^{2} – 2 times 5 times 7.4 times cos(38degree)\\
BC^{2} &= 25 + 54.76 – 58.312\\
BC^{2} &= 21.4472\\
BC &= sqrt{21.4472}\\
BC &approx 4.63 miles.\
end{aligned}$$
So, 4.63 miles far will Lisa have to hike from the lake to meet
Aaron at the campsite.

$=17^2+10^2-2cdot 17cdot 10cdot cos 45text{textdegree}$

$=389-340cdot 0.70710678$

$=148.5837$

$$
AB=sqrt{148.5837}approx 12.19
$$

$ABsin B=ACsin C$

$sin B=dfrac{ACsin C}{AB}=dfrac{10cdot 0.70710678}{12.19}$

$approx 0.5801$

$$
B=sin^{-1} 0.5801=35.5text{textdegree}
$$

$$
=dfrac{10cdot 17cdot 0.70710678}{2}approx 60.1
$$

b. Area $approx 60.1$

$$
boxed{a^2+b^2=c^2}
$$

$$
textit{*where $a$ and $b$ are length of legs and $c$ is length of hypotenuse of the right triangle}
$$

From the picture we can see this is a right triangle with both legs with the same length $x$ and length of hypotenuse is $7$. Now we can set the equation as:

$$
x^2+x^2=7^2
$$

$$
2x^2=49
$$

$$
x^2=dfrac{49}{2}
$$

$$
boxed{xapprox4.95}
$$

$$
boxed{sin theta=dfrac{a}{c}}
$$

$$
textit{*where $theta$ is angle $theta$, $a$ is length of opposite side, $c$ is length of hypotenuse of the right triangle}
$$

Since this is right triangle with legs which are same length, both angles will be $45$ degrees.

Now we can apply sine function as:

$$
sin 45text{textdegree}=dfrac{x}{7}
$$

$$
x=sin 45 text{textdegree} cdot 7
$$

$$
boxed{xapprox 4.95}
$$

$$
AB=BDsqrt 2=5sqrt 2approx 7.07
$$

$CD^2+5^2=10^2$

$CD^2=100-25=75$

$$
CD=sqrt{75}=5sqrt 3approx 8.66
$$

$=5sqrt 2+10+5+5sqrt 3$

$$
=7.07+10+13.66=30.73
$$

$=dfrac{(5+5sqrt 3)cdot 5}{2}=dfrac{25(1+sqrt 3)}{2}$

$$
=dfrac{13.66cdot 5}{2}=34.15
$$

Area: $dfrac{25(1+sqrt 3)}{2}approx34.15$

$$
i^5=i
$$

$$
i^6=-1
$$

$$
i^7=-i
$$

$$
i^8=1
$$

$$
i^9=i
$$

$$
i^{10}=-1
$$

$$
i^{11}=-i
$$

$$
i^{12}=1
$$

$$
i^{13}=i
$$

$$
i^{14}=-1
$$

$$
i^{15}=-i
$$

Thus we know that if $n$ is a multiple of 4, then $i^n=1$; if $n$ is one more than a multiple of 4, then $i^n=i$; if $n$ is 2 more than a multiple of 4, then $i^n=-1$. if $n$ is 3 more than a multiple of 4, then $i^n=-i$.

$$
i^{25}=i
$$

$$
i^{39}=-i
$$

$$
i^{100}=1
$$

c.
$$
i^{4n}=(i^4)^n=1^n=1
$$

d.
$$
i^{4n+1}=i^{4n}i=i
$$

$$
i^{4n+2}=i^{4n}i^2=-1
$$

$$
i^{4n+3}=i^{4n}i^3=-i
$$

e.
$$
i^{396}=1
$$

$$
i^{397}=i
$$

$$
i^{398}=-1
$$

$$
i^{399}=-i
$$

$r=0.035$

$A(t)=400,000(1+0.035)^t$

$$
A(t)=400,000(1.035)^t
$$

$400,000(1.035)^t=800,000$

$(1.035)^t=dfrac{800,000}{400,000}$

$(1.035)^t=2$

$log_{10} (1.035)^t=log_{10} 2$

$tlog_{10} 1.035=log_{10} 2$

$$
t=dfrac{log_{10} 2}{log_{10} 1.035}approx 20
$$

$$
P=200,000
$$

$A(t)=200,000(1-0.02)^t$

$$
A(t)=200,000(0.98)^t
$$

$5x-3y=12$          (Given)

$3y=5x-12$

$$
y=dfrac {5x-12}{3}
$$

$F=dfrac {Gm_{1}m_{2}}{r^2}$          (Given)

$Fr^2=Gm_{1}m_{2}$

$$
m_{2}=dfrac {Fr^2}{Gm_{1}}
$$

$E=dfrac {1}{2}mv^2$          (Given)

$2E=mv^2$

$$
m=dfrac {2E}{v^2}
$$

$(x-4)^2+(y-1)^2=10$          (Given)

$(y-1)^2=10-(x-4)^2$

$y-1=pm sqrt {10-(x-4)^2}$

$$
y=1 pm sqrt {10-(x-4)^2}
$$

b-          $F=dfrac {Gm_{1}m_{2}}{r^2} qquad rightarrow qquad m_{2}=dfrac {Fr^2}{Gm_{1}}$

c-          $E=dfrac {1}{2}mv^2 qquad rightarrow qquad m=dfrac {2E}{v^2}$

d-          $(x-4)^2+(y-1)^2=10 qquad rightarrow qquad y=1 pm sqrt {10-(x-4)^2}$

$$
begin{align*}
&dfrac{3}{x}+dfrac{2}{x+1}=5 tag{Given}\ \
&dfrac{3(x+1)+2x}{x(x+1)}=5 tag{Taking LCM}\ \
&3x+3+2x=5x(x+1) tag{Cross-multiplying}\ \
&5x+3=5x^2+5x tag{Adding and multiplying}\ \
&5x^2=3 tag{Simplifying}\ \
&x^2=dfrac{3}{5} tag{Dividing both side by 5}\ \
&x=pmsqrt{dfrac{3}{5}}tag{Solution}
end{align*}
$$

$$
begin{align*}
&x^2+6x+9=2x^2+3x+5 tag{Given}\ \
&6x+9=x^2+3x+5tag{Subtracting $x^2$ from both side}\ \
&x^2-3x-4=0 tag{Subtracting $6x$ $&$ 9 from both side}\ \
&x^2-4x+x-4=0 tag{$-3x$ as sum of $-4x+x$}\ \
&(x-4)(x+1)=0 tag{Factorising}\ \
&x=4 &x=-1 tag{Solution}\
end{align*}
$$

$$
begin{align*}
&8-sqrt{9-2x}=x+3 tag{Given}\ \
&-sqrt{9-2x}=x-5 tag{Simplifying}\ \
&9-2x=x^2-10x+25 tag{Squaring on both side}\ \
&x^2-8x+16=0 tag{Simplifying}\ \
&x^2-4x-4x+16=0 tag{$-8x$ as sum of $-4x & -4x$ }\ \
&(x-4)(x-4)=0 tag{Factorising}\ \
&x=4 tag{Solution}\
end{align*}
$$

$cos 30text{textdegree}=dfrac{5}{x}$

$xcdot dfrac{sqrt 3}{2}=5$

$xsqrt 3=5(2)$

$$
x=dfrac{10}{sqrt 3}=dfrac{10sqrt 3}{3}approx 5.77
$$

$$
log_5 (2x) = 3
$$

Apply $log$ rules:

$$
begin{align*}
2x &= 5^3 \
2x &= 125 \
x &= dfrac{125}{2}
end{align*}
$$

$$
begin{align*}
5^{-1} &= x+1 \
dfrac{1}{5} &= x+1 \
&text{Isolate $x$} \
x &= dfrac{1}{5} -1 \
x&= dfrac{-4}{5}
end{align*}
$$

$$
begin{align*}
log dfrac{4}{x} &= 2 \
dfrac{4}{x} &= 10^2 \
dfrac{4}{x} &= 100 \
100 x &= 4 \
&text{Isolate $x$:} \
x&= dfrac{4}{100} = 0.04
end{align*}
$$

$$
begin{align*}
log_3 6^2 + log_3 y &= 4 \
log_3 36y &= 4 \
3^4 &= 36 y \
81 &= 36y \
&text{Isolate $y$:} \
y&= dfrac{81}{36} \
y&= 2.25
end{align*}
$$