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elena-14-01-66 [18.8K]
3 years ago
12

A ball is thrown from a height of 157 feet with an initial downward velocity of 9/fts . The ball's height h (in feet) after t se

conds is given by the following. =h−157−9t-16tsquared How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth. (If there is more than one answer, use the "or" button.)

Mathematics
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer:

  2.86 seconds

Step-by-step explanation:

We presume the height is measured above the ground, so the height when the ball hits the ground is 0 feet. Putting that into the equation and solving for t, we get ...

  0 = 157 -9t -16t^2

We can use the quadratic formula with a=-16, b=-9, c=157 to find the solution.

  t = (-b -√(b^2 -4ac))/(2a) . . . . . we are not interested in the negative solution

  t = (-(-9) -√((-9)^2 -4(-16)(157)))/(2(-16))

  t = (9 -√10129)/-32 = (9 -100.6429)/-32 ≈ 2.86384

The ball will hit the ground about 2.86 seconds after it is thrown.

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At the zoo tje ratio of snakes to lizards is 3:2 A. If there were 10 lizards ,how many snakes would be there be? B.If there were
yawa3891 [41]

Answer:

Part A) 15\ snakes

Part B) 6\ lizards

Part C) The zoo would need to get four more lizards to maintain the same proportion

Part D) 12 snakes and 8 lizards

Step-by-step explanation:

Part A) If there were 10 lizards ,how many snakes would be there be?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2} ----> equation A

we have that

y=10\ lizards

substitute the value of y in the equation A

\frac{x}{10}=\frac{3}{2}

solve for x

x=10(3)/2\\x=15\ snakes

Part B) If there were 9 snakes ,how many lizards would there be?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2} ----> equation A

we have that

x=9\ snakes

substitute the value of x in the equation A

\frac{9}{y}=\frac{3}{2}

solve for y

y=9(2)/3\\y=6\ lizards

Part C) If the number of snakes in the zoo is increased by 6, how many more lizards would the zoo need to get to keep the same ratio?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2} ----> equation A

For x=6\ snakes

substitute the value of x in the equation A

\frac{6}{y}=\frac{3}{2}

solve for y

y=6(2)/3\\y=4\ lizards

therefore

The zoo would need to get four more lizards to maintain the same proportion

Part D) If the total number of snakes and lizards at the zoo was 20, how many snakes and lizards would there be?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2}

isolate the variable x

x=1.5y ----> equation A

x+y=20 ----> equation B

solve the system by substitution

substitute equation A in equation B

1.5y+y=20

solve for y

2.5y=20

y=8\ lizards

Find the value of x

x=1.5(8)=12\ snakes

therefore

12 snakes and 8 lizards

8 0
3 years ago
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