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jeka94
3 years ago
15

You work 4 hours on Saturday and 8 hours on Sunday. You also receive a $50 bonus. You earn $164. How much did you earn per hour?

Mathematics
2 answers:
Vesna [10]3 years ago
5 0
You earned 13.6 per hr
Marina86 [1]3 years ago
3 0
To find the hourly rate, we would first need to take the total earnings and subtract the bonus. $164 - $50 = 114 This gives us the total money earned for hourly work. Then we need to add the total hours worked: 4 + 8 = 12. Lastly, we need to take the total money earned and divide it by the total hours worked: 114 / 12 = 9.50 or $9.50 per hour. The equation could look like: 4h + 8h + $50 = $165
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What would be the dimensions for the poster at 1/4 times it’s current size
iVinArrow [24]

Answer:

D. Length = 9 cm; width = 6 cm

Step-by-step explanation:

Find 1/4 of the length and 1/4 of the width.

1/4 * 36 cm = 9 cm

1/4 * 24 cm = 6 cm

Answer: D. Length = 9 cm; width = 6 cm

6 0
3 years ago
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Nataly_w [17]

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The average rate of change is 4.

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2 years ago
Please anyone answer me
ollegr [7]

Let's divide the shaded region into two areas:

area 1: x = 0 ---> x = 2

ares 2: x = 2 ---> x = 4

In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.

Area 1:

A\frac{}{1}  = ∫g(x)dx =  ∫xdx =  \frac{1}{2}  {x}^{2}  | \frac{2}{0}  = 2

Area 2:

A\frac{}{2}  = ∫(g(x) - f(x))dx

= ∫(x -  {(x - 2)}^{2} )dx

=  ∫( - {x}^{2}  + 5x - 4)dx

= ( - \frac{1}{3}{x}^{3} +   \frac{5}{2} {x}^{2}  - 4x)   | \frac{4}{2}

= 2.67 - ( - 0.67) = 3.34

Therefore, the area of the shaded portion of the graph is

A = A1 + A2 = 5.34

3 0
3 years ago
Please help me for the 2nd part
Firlakuza [10]

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6 0
2 years ago
Enter the correct answer in the box. What is the standard form polynomial that represents this product?
JulsSmile [24]

Answer: -8m^5+10m^4+9m^3-16m^2+5m

Step-by-step explanation:

I had this question on a recent test and got it right

Hope this helps! :)

8 0
3 years ago
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