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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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1.)\\ \\ b^2+10b=75\\ \\ b^2+10b-75=0\\ \\\Delta = b^{2}-4ac = 10^{2}-4*1*(75)=100+300=400 \\ \\\sqrt{\Delta }=\sqrt{400}=20

b_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-10-20}{2}=\frac{-30}{2}=-15\\ \\b_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{-10+20}{2}=\frac{10}{2}=5



2.)\\ \\ n^2-20n=-75\\ \\ n^2-20n+75=0\\ \\\Delta = b^{2}-4ac = (-20)^{2}-4*1*75=400-300=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10

n_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{20-10}{2}=\frac{10}{2}=5\\ \\n_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{20+10}{2}=\frac{30}{2}=15



3.)\\ \\t^2+8t-9=0\\ \\\Delta = b^{2}-4ac = 8^{2}-4*1* (-9)=64+36=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10

t_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-8-10}{2}=\frac{-18}{2}=-9\\ \\t_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{-8+10}{2}=\frac{2}{2}=1


4.)\\ \\m^2-2m-8=0\\ \\\Delta = b^{2}-4ac = (-2)^{2}-4*1* (-8)=4+32=36\\ \\\sqrt{\Delta }=\sqrt{36}=6

m_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{2-6}{2}=\frac{-4}{2}=-2\\ \\m_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{2+6}{2}=\frac{8}{2}=4



5.)\\ \\ v^2+4v-2=0\\ \\\Delta = b^{2}-4ac = 4^{2}-4*1* (-2)=16+8=24\\ \\\sqrt{\Delta }=\sqrt{24}= \sqrt{4*6}=2\sqrt{6}

v_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-4-2\sqrt{6}}{2}=\frac{2(-2-\sqrt{6}}{2}= -2-\sqrt{6}\approx -2-2,45\approx -4,45\\ \\v_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{-4+2\sqrt{6}}{2}=\frac{2(\sqrt{6}-2}{2}= \sqrt{6}-2 \approx 2,45-2\approx 0,45


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