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bazaltina [42]
3 years ago
12

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

e given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+128x+136
Mathematics
1 answer:
Anni [7]3 years ago
7 0

Answer: The rocket will reach its max after 4 seconds

Step-by-step explanation:

Hi, to find the maximum height, we have to set the first derivative of the equation equal to zero.

y=-16x^2+128x+136

0 = (2)-16x +128  

0= -32x+128

Solving for x:

32x = 128

x = 128/32

x = 4 seconds

The rocket will reach its max after 4 seconds

Feel free to ask for more if needed or if you did not understand something.

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E varies directly with G. When G is 6, E is 10. Find the value of G when E is 18.
sasho [114]

Answer:

  10.8

Step-by-step explanation:

E has increased by a factor of 18/10 = 1.8, so G will increase by that factor. The corresponding value of G is ...

  6 × 1.8 = 10.8 = G

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If you like, you can write an equation relating G to E:

  E = kG

  10 = k(6)

  k = 10/6 = 5/3

Now for E = 18, we have ...

  18 = (5/3)G

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7 0
3 years ago
if the HCF of 408 and 1032 is expressible in the form of 1032 into 2 + 408 into P then find the value of p​
alexandr1967 [171]

Answer:

below.

Step-by-step explanation:

408 = 2*2*2*3*17

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So the HCF is 2*2*2*3 = 24

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24 - 2/1032 = p/408

p = (24 - 2/1032) * 408

p = 9791.21 to nearest hundredth.

3 0
3 years ago
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