$f(x)=sqrt {x+4}$          (Given)

$g(x)=x^2-x$          (Given)

a-          $f(5)=sqrt {5+4}=sqrt {9}=3$

b-          $g(-1)=(-1)^2-(-1)=1+1=2$

c-          $f(x)=10$

$10=sqrt {x+4}$

$100=x+4$

$$
x=96
$$

d-          $g(x)=6$

$6=x^2-x$

$x^2-x-6=0$

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

$x-3=0$ or $x+2=0$

$$
x=3 textrm { ~or ~} x=-2
$$

a-          $f(5)=3$

b-          $g(-1)=2$

c-          $x=96$

d-          $x=3$ or $x=-2$

(a)
$\\$The x-values are less than 6 therefore the domain is$$text{color{#4257b2}(-infty, 6)$.
$\\$The y-values are less than 6 therefore the range is $color{#4257b2}(-infty, 6)$.}$

(b)
$\\$All x-values are part of the domain therefore the domain is$$text{color{#4257b2}(-infty, +infty)$.
$\\$The y-values are from -3 to 3 therefore the range is $color{#4257b2}[-3, 3]$.}$

a-          $y=-dfrac {3}{2}x+8$

x-intercept is $(5.33,0)$

y-intercept is $(0.8)$

b-          $2x-3y=-6$

x-intercept is $(-3, 0)$

y-intercept is $(0.2)$

a-          $y=-dfrac {3}{2}x+8$

x-intercept is $(5.33,0)$

y-intercept is $(0.8)$

b-          $2x-3y=-6$

x-intercept is $(-3, 0)$

y-intercept is $(0.2)$

a)

$$
begin{equation}
frac{x+2}{5}=frac{10-2x}{3}
end{equation}
$$

We can multiply the whole equation by $15$ to get rid of the fractions.

$$
begin{equation}
frac{15(x+2)}{5}=frac{15(10-2x)}{3}
end{equation}
$$

Now we can divide $15$ by $5$ on the left hand side and $15$ by $3$ on the right hand side of the equation.

$$
begin{equation}
3(x+2)=5(10-2x)
end{equation}
$$

If we multiply each side of the equation (3) we get:

$$
begin{equation}
3x+6=50-10x
end{equation}
$$

Now we can separate parts of the equation (4) with $x$s and those without.

$$
begin{equation}
3x+10x=50-6
end{equation}
$$

If we sum that we get:

$$
begin{equation}
13x=44
end{equation}
$$

Now if we divide equation (6) by $13$ we get the value of $x$.

$$
begin{equation}
x=frac{44}{13}
end{equation}
$$

b)

$$
begin{equation}
frac{3}{x}-1=8
end{equation}
$$

We can start by moving $-1$ to the right hand side of the equation.

$$
begin{equation}
frac{3}{x}=8+1
end{equation}
$$

$$
begin{equation}
frac{3}{x}=9
end{equation}
$$

We can also divide the whole equation (3) by $3$.

$$
begin{equation}
frac{1}{x}=3
end{equation}
$$

If we now use the reciprocal equation of equation (4) we have the value of $x$.

$$
begin{equation}
x=frac{1}{3}
end{equation}
$$

c)

$$
begin{equation}
x^2+3x=18
end{equation}
$$

We can start solving the equation by moving $18$ to the left hand side of the equation (1). That way we can solve equation (1) using the quadratic formula.

$$
begin{equation}
x^2+3x-18=0
end{equation}
$$

Quadratic formula:

$$
begin{equation}
x_{1,2}=frac{-bpmsqrt{b^2-4ac}}{2a}
end{equation}
$$

For this equation $a=1$, $b=3$ and $c=-18$.
So now we can write:

$$
begin{equation}
x_{1,2}=frac{-3pmsqrt{3^2-4cdot1cdot(-18)}}{2cdot1}
end{equation}
$$

$$
begin{equation}
x_{1,2}=frac{-3pmsqrt{9+72}}{2}
end{equation}
$$

$$
begin{equation}
x_{1,2}=frac{-3pmsqrt{81}}{2}
end{equation}
$$

$$
begin{equation}
x_{1,2}=frac{-3pm9}{2}
end{equation}
$$

If we separate the two equations:

$$
begin{equation}
x_1=frac{-3+9}{2}
end{equation}
$$

$$
begin{equation}
x_2=frac{-3-9}{2}
end{equation}
$$

From equation (8) $x_1=3$ and from equation (9) $x_2=-6$.

a) $x=frac{44}{13}$

b) $x=frac{1}{3}$

c) $x_1=3$, $x_2=-6$

$f(x)=x^2-2x-8$          (Given)

The domain of the function is:          $(-infty<x<infty)$

The range of the function is:          $(-9<y<infty)$

x-intercepts are: $(-2, 0)$ and $(4, 0)$

y-intercepts is: $(0, -8)$

The vertex of the parabola is:          $(1, -9)$

Domain:          $(-infty<x<infty)$                   Range:          $(-9<y<infty)$

x-intercepts are: $(-2, 0)$ and $(4, 0)$                   y-intercepts is: $(0, -8)$

The vertex of the parabola is:          $(1, -9)$

(a)
$\\$The graphs will intersect at the point where the two equations have the same y-values.$

$To find this point, equate the two equations to each other to have:$

3x+15 = 3-3x

$Solve for$x$to have:$

3x+3x=3-15

6x=-12

x = $frac{-12}{6}$

x=-2

$Find the value of$y$when$x = -2$to have:$

y=3x+15

y=3(-2)+15

y=-6+15

y=9

$Thus, the graphs will intersect at the point$(-2, 9)$.

(b)
$\\$The graphs will intersect at the point/s where the y-value of$y=x^2-3x-8$is equal to 2.$

$To find this point, substitute 2 to the$y$of the first equation to obtain:$

y= x^2-6x-8

2 = x^2-3x-8

0 = x^2-3x-8-2

0=x^2-3x-10

$Factor the trinomial to obtain:$

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

$Equate each factor to 0 then solve each equation to have:$

x-5=0 or x + 2 =0

x=5 or x = -2

$Thus, the graph will intersect in the points$(-2, 2)$and$(5, 2)$.

a.

Let us solve the following equation
$$
color{#4257b2} x^2-x-6=0.
$$

Since the expression of the left is a quadratic function we can use the quadratic formula

$$
x = frac{-b pm sqrt{b^2-4ac}}{2a}.
$$
We have

$$
a=1 hspace{12mm} b=-1 hspace{12mm} c=-6.
$$

Hence, we have

$$
begin{align*}
x &= frac{-(-1) pm sqrt{(-1)^2-4 cdot 1 cdot (-6)}}{2 cdot 1} \
x &= frac{1 pm sqrt{1+24}}{2} \
x &= frac{1 pm sqrt{25}}{2} \
x &= frac{1 + sqrt{25}}{2} hspace{3mm} text{or} hspace{3mm} frac{1 – sqrt{25}}{2} \
x &= frac{1 + 5}{2} hspace{3mm} text{or} hspace{3mm} frac{1 – 5}{2} \
x &= frac{6}{2} hspace{3mm} text{or} hspace{3mm} frac{-4}{2} \
x &=3 hspace{3mm} text{or} hspace{3mm} -2
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x=3 hspace{3mm} text{color{default}or} hspace{3mm} x=-2.
$$

b.

Let us solve the following equation
$$
color{#4257b2} 5x^2-8=12x.
$$
First we transform the equation a little bit and get

$$
5x^2-12x-8=0.
$$

Since the expression of the left is a quadratic function we can use the quadratic formula

$$
x = frac{-b pm sqrt{b^2-4ac}}{2a}.
$$
We have

$$
a=5 hspace{12mm} b=-12 hspace{12mm} c=-8.
$$

Hence, we have

$$
begin{align*}
x &= frac{-(-12) pm sqrt{(-12)^2-4 cdot 5 cdot (-8)}}{2 cdot 5} \
x &= frac{12 pm sqrt{144+160}}{10} \
x &= frac{12 pm sqrt{304}}{10} \
x &= frac{12 + sqrt{304}}{10} approx 2,94 hspace{3mm} text{or} hspace{3mm} frac{12 – sqrt{304}}{10} approx -0,54.
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x=frac{12 + sqrt{304}}{10} approx 2,94 hspace{3mm} text{color{default}or} hspace{3mm} x=frac{12 – sqrt{304}}{10} approx -0,54.
$$

c.

Let us solve the following equation
$$
color{#4257b2} x^2-8x-20=0.
$$

Since the expression of the left is a quadratic function we can use the quadratic formula

$$
x = frac{-b pm sqrt{b^2-4ac}}{2a}.
$$
We have

$$
a=1 hspace{12mm} b=-8 hspace{12mm} c=-20.
$$

Hence, we have

$$
begin{align*}
x &= frac{-(-8) pm sqrt{(-8)^2-4 cdot 1 cdot (-20)}}{2 cdot 1} \
x &= frac{8 pm sqrt{64+80}}{2} \
x &= frac{8 pm sqrt{144}}{2} \
x &= frac{8 + sqrt{144}}{2} hspace{3mm} text{or} hspace{3mm} frac{8 – sqrt{144}}{2} \
x &= frac{8 + 12}{2} hspace{3mm} text{or} hspace{3mm} frac{8 – 12}{2} \
x &= frac{20}{2} hspace{3mm} text{or} hspace{3mm} frac{-4}{2} \
x &=10 hspace{3mm} text{or} hspace{3mm} -2
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x=10 hspace{3mm} text{color{default}or} hspace{3mm} x=-2.
$$

d.

Let us solve the following equation
$$
color{#4257b2} 2y^2-5y=12.
$$
First we transform the equation a little bit and get

$$
2y^2-5y-12=0.
$$

Since the expression of the left is a quadratic function we can use the quadratic formula

$$
y = frac{-b pm sqrt{b^2-4ac}}{2a}.
$$
We have

$$
a=2 hspace{12mm} b=-5 hspace{12mm} c=-12.
$$

Hence, we have

$$
begin{align*}
y &= frac{-(-5) pm sqrt{(-5)^2-4 cdot 2 cdot (-12)}}{2 cdot 2} \
y &= frac{5 pm sqrt{25+96}}{4} \
y &= frac{5 pm sqrt{121}}{4} \
y &= frac{5 + 11}{4} hspace{3mm} text{or} hspace{3mm} frac{5 – 11}{4} \
y &= frac{16}{4} hspace{3mm} text{or} hspace{3mm} frac{-6}{4} \
y &= 4 hspace{3mm} text{or} hspace{3mm} -1,5.
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} y=4 hspace{3mm} text{color{default}or} hspace{3mm} y=-1,5.
$$

$$
begin{align*}
&text{a.} hspace{3mm} x=3 hspace{3mm} text{color{default}or} hspace{3mm} x=-2 \
&text{b.} hspace{3mm} x=frac{12 + sqrt{304}}{10} approx 2,94 hspace{3mm} text{color{default}or} hspace{3mm} x=frac{12 – sqrt{304}}{10} approx -0,54 \
&text{c.} hspace{3mm} x=10 hspace{3mm} text{color{default}or} hspace{3mm} x=-2 \
&text{d.} hspace{3mm} y=4 hspace{3mm} text{color{default}or} hspace{3mm} y=-1,5 \
end{align*}
$$

The following is the $(x rightarrow y)$ table of the relation.\
begin{center}
renewcommand{arraystretch}{1.5}
begin{tabular}{|r|r|}
hline
Weeks $x$ & Remaining $y$ \
hline
1 & 185 \
hline
2 & 170 \
hline
3 & 155 \
hline
4 & 140 \
hline
5 & 125 \
hline
6 & 110 \
hline
7 & 95 \
hline
8 & 80 \
hline
9 & 65 \
hline
10 & 50 \
hline
11 & 35 \
hline
12 & 20 \
hline
13 & 5 \
hline
end{tabular}
end{center}
The initial value is $200\
The rate of change is $15 per week.\
The equation that represents this relation is:\
$$y=200-15x$$
\
When the money is gone $y=0$\
$0=200-15x$\
$x=dfrac {200}{15}$\
$x=13.3333$\
The money is gone after 13.3333 weeks.

The function that represents the relation is:          $y=200-15x$

The money is gone after 13.3333 weeks.