Question

1. the table shows the total wages a worker earned based on the number of hours they worked. hourly wage

number of hours	wages
15	$843.75
23	$1,293.75
30	$1,687.50

what is the average rate of change for the wages earned between 12 and 30 hours?

1. graph a line with a slope of -\frac{1}{4} that passes through the point (-4, 6).
2. place an x in the table to show whether each equation is linear or nonlinear.

equation	linear	nonlinear
27x - 13y = 42
4y + 2x = 3
9.25x^2 + 2.75y^2 = 35

1. a line contains the points (3, 5) and (11, 12).

part a
what is the slope of the line?
part b
graph the line.

Explanation:

Question 9

Step1: Recall average rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[x_1,x_2]$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Here, $x$ is the number of hours and $y$ is the wages. Let $x_1 = 12$, $y_1=675$, $x_2 = 30$, and $y_2 = 1687.50$.

Step2: Substitute values into formula

$\frac{y_2 - y_1}{x_2 - x_1}=\frac{1687.50 - 675}{30 - 12}$.
First, calculate the numerator: $1687.50−675 = 1012.5$.
Then, calculate the denominator: $30 - 12=18$.

Step3: Calculate the result

$\frac{1012.5}{18}=56.25$.

Step1: Recall slope formula

The slope $m$ of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here, $x_1 = 3$, $y_1 = 5$, $x_2 = 11$, and $y_2 = 12$.

Step2: Substitute values into formula

$m=\frac{12 - 5}{11 - 3}=\frac{7}{8}$.

Answer:

$56.25$

Question 10

The equation of a line in point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Given $m =-\frac{1}{4}$ and the point $(-4,6)$ (so $x_1=-4$ and $y_1 = 6$), the equation of the line is $y - 6=-\frac{1}{4}(x + 4)$.
To graph the line:

1. Plot the point $(-4,6)$.
2. Use the slope $m =-\frac{1}{4}$. The slope means that for every 4 units we move to the right (increase in $x$), we move 1 unit down (decrease in $y$). Starting from the point $(-4,6)$, if we move 4 units to the right to $x = 0$, then $y=6 - 1=5$. So we have another point $(0,5)$. We can plot more points in a similar fashion and draw a straight line through them.

Question 11

1. For the equation $y = 7(4^x)$:

• A linear equation is of the form $y=mx + b$ where $m$ and $b$ are constants and the highest power of $x$ is 1. In $y = 7(4^x)$, the variable $x$ is in the exponent. So it is a non - linear equation.

1. For the equation $27x-13y = 42$, we can rewrite it as $y=\frac{27}{13}x-\frac{42}{13}$, which is in the form $y = mx + b$. So it is a linear equation.
2. For the equation $4y+2x = 3$, we can rewrite it as $y=-\frac{1}{2}x+\frac{3}{4}$, which is in the form $y = mx + b$. So it is a linear equation.
3. For the equation $9.25x^{2}+2.75y^{2}=35$, the highest power of $x$ and $y$ is 2. So it is a non - linear equation.

Equation	Linear	Nonlinear
$27x-13y = 42$	X
$4y + 2x = 3$	X
$9.25x^{2}+2.75y^{2}=35$		X

Question 12

Part A