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kupik [55]
3 years ago
11

An airplane is traveling at 400 miles per hour. Which equation can be used to find the total distance the plane will travel in h

hours?
1. d = h ÷ 400

2. d = h + 400

3. d = 400 x h

4. d = 400 - h​
Mathematics
2 answers:
astra-53 [7]3 years ago
8 0

Answer:

Step-by-step explanation:

1

liq [111]3 years ago
3 0

Answer:

For this question, all we need to do is to apply the formular to calculate the distance since we already had the time which is h hours and the speed which is 400 miles per hour:

d = v × t = 400 × h

So the correct answer is number 3.

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NemiM [27]

Answer:

x = 55

Step-by-step explanation:

Draw a line parallel to the top and bottom parallel lines so this new line goes through the pointed end of x.

Draw another line parallel to the top and bottom lines through the pointy end of 45.

The bottom angle of the line through 45 is 15 degrees (alternate interior angles.

The top angle is 45 - 15 = 30

The bottom angle of the line going to x is 150 degrees. It and the 30 degree angle make 180. 30 + 150 = 180

One final observation The top angle of made by the line going through x is 180 - 25 = 155

What you have now is

155 + 150 + x = 360

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Step-by-step explanation:

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712 ÷ 74 is equal to _____.<br><br> 3<br> 7 3<br> 1 3<br> 7 8
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Answer:

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A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 w
Liono4ka [1.6K]

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds.  Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

t = 5.0625

Step-by-step explanation:

Given

h(t)=-16t^2+80t+5

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

0=-16t^2+80t+5

Reorder

-16t^2+80t+5 = 0

Using quadratic formula, we have:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

Where

a = -16      b =80      c = 5

So, we have:

t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}

t = \frac{-80\±\sqrt{6400 +320}}{-32}

t = \frac{-80\±\sqrt{6720}}{-32}

t = \frac{-80\±82.0}{-32}

This gives:

t = \frac{-80+82.0}{-32} or t = \frac{-80-82.0}{-32}

t = \frac{2}{-32} or t = \frac{-162}{-32}

t = -\frac{2}{32} or t = \frac{162}{32}

But time can not be negative.

So, we have:

t = \frac{162}{32}

t = 5.0625

<em>Hence, time to hit the ground is 5.0625 seconds</em>

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