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soldi70 [24.7K]
3 years ago
8

Joshua earned $60 for working h hours. Which expression represents how much he earned in 1 hour?

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
6 0

we know that

Joshua earned \$60 for working h hours

so

by proportion

Find how much he earned in 1 hour

\frac{60}{h} \frac{\$}{hours}=\frac{x}{1} \frac{\$}{hour}\\ \\x=\$(\frac{60}{h})

therefore

<u>the answer is the option D</u>

(\frac{60}{h})

Shkiper50 [21]3 years ago
5 0
D. 60 divided by h.

Hope this helped.
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zalisa [80]
The answer is Option B. It would be cheaper for Miguel to drive 200 miles with Option B than Option A.
8 0
4 years ago
Read 2 more answers
What is the equation in vertex form of the quadratic function with a vertex at (-1, -4) that goes through (1, 8)?
cestrela7 [59]

Answer:

y = 3(x+1)^2 - 4

Step-by-step explanation:The general form of the equation of a quadratic function whose vertex is (h,k) and whose leading coefficient is a is:

y - k = a(x-h)^2, or

y      = a(x-h)^2 - k

Substituting the coefficients of the vertex (-1, -4), we get:

y      = a(x + 1)^2 - 4

Substituting the coordinates of the given point, (1,8), we get:

8      = a(1+1)^2 - 4, which simplifies to:

8      = a(2)^2 - 4, or

8  = 4a - 4.  Then 4a = 12, and a = 3.

Thus, the desired equation is y = 3(x+1)^2 - 4 (answer j).


5 0
4 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
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