$bold{a.)}$ In this case,

$m = -dfrac{2}{5}$

$b = 8$

Using the slope-intercept form

$$
y= -dfrac{2}{5}x+8
$$

Remember the $bold{slope}$ $bold{-intercept ;form}$ of a line:

$y=mx+b$

where

$m$ = slope

$b$ = y-intercept

$bold{b.)}$ We are given with,

$m=-9$

point on the line = $(5,4)$

Using the point-slope form,

$y=-9(x-5)+4$

$y=-9x-45+4$

$$
y=-9x+49
$$

The $bold{point}$ $-$ $bold{slope; form}$ of a line is

$y=m(x-x_1)+y_1$

where

$m = slope$

$(x_1,y_1)$ = any point on the line

$bold{c.)}$ The points are

$(5,0)$ and $(0,-7)$

$m=dfrac{-7-0}{0-5}=dfrac{7}{5}$

The point-slope form is written as

$y=m(x-x_1)+y_1$

$y=dfrac{7}{5}(x-5)+0$

$$
y=dfrac{7}{5}x-7
$$

If given two points $(x_1,y_1)$ and $(x_2,y_2)$, then the slope of the line is

$$
m=dfrac{y_2-y_1}{x_2-x_1}
$$

$bold{d.)}$ The given points are

$(-12,17)$ and $(5,-17)$

$m=dfrac{-17-17}{5-(-12)}=-dfrac{34}{17}=-2$

Thus, the point-slope form is

$y=-2(x+12)+17$

$y=-2x-24+17$

$$
y=-2x-7
$$

a.) $y=-dfrac{2}{5}x+8$

b.) $y=-9x+49$

c.) $y=dfrac{7}{5}x-7$

d.) $y=-2x-7$

$bold{a.)}$ In this case,

$g(-2)=3(-2)+(-2)-4$

$$
g(-2)=6
$$

Given a function $f(x)$,
to evaluate the function at $x=a$,
simply substitute the value of $a$ to x.

$bold{b.)}$

$fleft(dfrac{3}{4}right)=dfrac{4}{5}cdot dfrac{3}{4}+dfrac{7}{10}$

$fleft(dfrac{3}{4}right)=dfrac{3}{5}+dfrac{7}{10}$

$$
fleft(dfrac{3}{4}right)=dfrac{13}{10}
$$

Do similar procedure for parts $b$ to $d$.

Remember that

$$
dfrac{a}{b}cdot dfrac{c}{a}=dfrac{c}{b}
$$

$bold{c.)}$

$g(sqrt{6})=4(sqrt{6})^2-24$

$g(sqrt{6})=4(6)-24$

$$
g(sqrt{6})=0
$$

Remember that $(sqrt{a})^2=a$

$bold{d.)}$

$hleft(dfrac{1}{3}right)=64^{1/3}$

$hleft(dfrac{1}{3}right)=(4^3)^{1/3}$

$$
hleft(dfrac{1}{3}right)=4
$$

Remember that

$$
(a^b)^{1/b}=a
$$

a.) 6

b.) $dfrac{13}{10}$

c.) 0

d.) 4

To determine whether the relation is linear, quadratic, or neither linear not quadratic, remember the following.

(1) If the 1st differences are constant, the relation is $bold{linear}$.

(2) If the 1st differences are not constant, but the 2nd differences are constant, then it is $bold{quadratic}$.

(3) If neither of the above statement is true, then it is $bold{neither; linear; nor; quadratic}$.

a.) The first differences are $5,5,5,5$

Thus, it is $bold{linear}$.

b.) The first differences are $6,12,24,18$.

$therefore$ not linear

The second differences are:

$12-6=6$

$24-12=12$

$48-24=24$

$therefore$ not quadratic

Thus, the relation is $bold{neither;linear;nor;quadratic}$.

c.) The first differences are $2,6,10,14$

$therefore$ not linear

The second differences are

$6-2=4$

$10-6=4$

$14-10=4$

$therefore$ quadratic

Thus, the relation is $bold{quadratic}$.

$bold{a.)}$ $2x-$3 = 7

$2x=7+3$

$2x=10$

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

$$
x=5
$$

The general principle is to group terms with a variable on one side.

Remember that

$$
ax=bimplies x=dfrac{b}{a}
$$

$bold{b.)}$ $5x+8=2x-7$

$5x-2x=-7-8$

$3x=-15$

$dfrac{3x}{3}=dfrac{-15}{3}$

$$
x=-5
$$

$bold{c.)}$

$5(3x-2)+7x-4=2(4x+8)-2x+3$

$15x-10+7x-4=8x+16-2x+3$

$22x-14=6x+19$

$22x-6x=19+14$

$16x=33$

$dfrac{16x}{16}=dfrac{33}{16}$

$$
x=dfrac{33}{16}
$$

Remember the distributive property

$$
a(x+c)=ax+ac
$$

$bold{d.)}$

$-8x+dfrac{3}{4}=dfrac{1}{3}x-12$

$-8x-dfrac{1}{3}x=-12-dfrac{3}{4}$

$-dfrac{25}{3}x=-dfrac{51}{4}$

$x=-dfrac{3}{25}cdot dfrac{-51}{4}$

$$
x=dfrac{153}{100}
$$

Remember that

$$
dfrac{a}{b}x=cimplies x=dfrac{b}{a}cdot c
$$

a.) $x=5$

b.) $x=-5$

c.) $x=dfrac{33}{16}$

d.) $x=dfrac{153}{100}$

$bold{Half}$ $bold{-life}$ $(t_{1/2})$ is the time to reach half of the original amount. If the starting amount is $M_0$, the amount remaining after time $t$ which we will denote as $M(t)$ is calculated as

$$
M(t)=M_0left(dfrac{1}{2}right)^{t/t_{1/2}}
$$

In this case, $t=1000$ years, $t_{1/2}=100$ years,

$M(1000)=2.3left(dfrac{1}{2}right)^{1000/10}$

$M(1000)=2.3left(dfrac{1}{2}right)^{10}=dfrac{2.3}{1024}=0.022$ kg

$0.022$ kg

$bold{Doubling}$ $bold{-time}$ $(t_{2})$ is the time to reach double of the original amount. If the starting amount is $M_0$, the amount remaining after time $t$ which we will denote as $M(t)$ is calculated as

$$
M(t)=M_0left( 2 right)^{t/t_{2}}
$$

In this case, $t=9$ weeks, $t_{2}=1$ week,

$M(9)=0.1left(2right)^{9/1}$

$$
M(9)=0.512=51.2%
$$

$$
51.2%
$$

$bold{Definition}$

Exponential function is a function such that the independent variable is in the exponent which can be written in the form

$y=acdot b^{(ccdot x)}$

where $a,;b,$ and $c$ are constants.

$bold{Methods}$

For an exponential function $y=acdot b^{ccdot x}$

domain: $bold{R}$

range:
if $a>0$, ${ yin bold{R};|;y>0}$

if $a<0$ , ${ yinbold{R};|;y<0}$

$y$-intercept: $(0,a)$

$bold{Examples}$

$y=2^x$

$f(x)=0.5^{3x}$

$$
g(t)=dfrac{1}{3^{0.4t}}
$$

$bold{Non-examples}$

$y=xcdot 3^x$

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

$y=2x+3$

$y=x^2+2x+1$