Evaluate and explain how you evaluated 16^3/2

Joe's annual income has been increasing each year by the same dollar amount. The first year his income was $17 comma 90017,900

Vedmedyk [2.9K]

Answer:

In 17th year, his income was $30,700.

Step-by-step explanation:

It is given that the income has been increasing each year by the same dollar amount. It means it is linear function.

Income in first year = $17,900

Income in 4th year = $20,300

Let y be the income at x year.

It means the line passes through the point (1,17900) and (4,20300).

If a line passes through two points $(x_1,y_1)$ and $(x_2,y_2)$ , then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of line is

$y-17900=\frac{20300-17900}{4-1}(x-1)$

$y-17900=\frac{2400}{3}(x-1)$

$y-17900=800(x-1)$

$y-17900=800x-800$

Add 17900 on both sides.

$y=800x-800+17900$

$y=800x+17100$

The income equation is y=800x+17100.

Substitute y=30,700 in the above equation.

$30700=800x+17100$

Subtract 17100 from both sides.

$30700-17100=800x$

$13600=800x$

Divide both sides by 800.

$\frac{13600}{800}=x$

$17=x$

Therefore, in 17th year his income was $30,700.

5 0

3 years ago

Q7) A rectangular piece of land measures 8 m by 5m. It is to be fenced with four lines of wire. What length of wire is needed fo

Harrizon [31]

Answer:

26 m

Step-by-step explanation:

Find what length is needed for fencing by finding the perimeter of the land.

Use the perimeter formula, P = 2l + 2w

Plug in the length and width:

P = 2l + 2w

P = 2(8) + 2(5)

P = 16 + 10

= 26

So, the length of wire needed is 26 m

8 0

2 years ago

Please Help I am kinda confused I will give Brainliest.

Illusion [34]

Answer:

C

Step-by-step explanation:

its c

4 0

3 years ago

Read 2 more answers

Suppose that a function f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence

Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

$\lim_{x \to c} \frac{x^2-cx}{(x-c)}$

We can simplify it

$\lim_{x \to c} \frac{x(x-c)}{(x-c)}$

$=\lim_{x \to c} x$

$=c$

so, we can see that limit exist and it's value defined

case-2:

$\lim_{x \to c} \frac{1}{(x-c)}$

Left limit is

$\lim_{x \to c-} \frac{1}{(x-c)}$

$=-\infty$

Right Limit is

$\lim_{x \to c+} \frac{1}{(x-c)}$

$=+\infty$

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0

3 years ago

Select the situation that can be represented by 20x

ipn [44]

You work x hours. you get 20 dollars per hour. this is a random situation, because u didn't give us choices.

3 0

3 years ago