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MArishka [77]
3 years ago
10

An object is thrown upward at a speed of 152 feet per second by a machine from a height of 9 feet off the ground. The height h o

f the object after t seconds can be found using the equation h=−16t^2+152t+9
When will the height be 231 feet?


When will the object reach the ground?

Mathematics
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

First part:

Set h(t) = 231and solve for t.

-16t²+ 152t + 9= 231

-16t² + 152t - 222= 0

Solve this quadratic equation for t. You should get 2 positive solutions. The lower value is the time to reach 231 on the way up, and the higher value is the time to reach 231 again, on the way down.

Second part:

Set h(t) = 0 and solve the resulting quadratic equation for t. You should get a negative solution (which you can discard), and a positive solution. The latter is your answer.

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The volume of a cylinder is 252π cm3 and its height is 7 cm. What is the cylinder’s radius? A. 36 B. 80 C.16 D.6
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3 years ago
a triangle with vertices at A (-1,1), B(-2,1) C(-1,4) is translated. the image of vertex A has coordinate at (3,-1) determine th
Drupady [299]
For this case we have the following vertices:
 A (-1,1) --------> A '(3, -1)
 B (-2.1)
 C (-1,4)
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4 years ago
Solve for y<br> 2(3 + 3y) + y = 11
garik1379 [7]
<h2><u>EQUATION</u></h2><h3>Exercise</h3>

2(3 + 3y) + y = 11

First, apply the distributive property:

2(3 + 3y) + y = 11

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6 + 7y = 11

Substract 6 from both sides:

6 - 6 + 7y = 11 - 6

7y = 5

Divide both sides by 7:

\dfrac{7y}{7} = \dfrac{5}{7}

\boxed{y = \dfrac{5}{7}}

<h3><u>Answer</u>. The value of y = 5/7.</h3>

8 0
3 years ago
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