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taurus [48]
2 years ago
6

The height in feet h of a water bottle rocket launched from a rooftop is given by the equation h = -16t2 + 320t + 32, where t is

the time in seconds. After the rocket is fired, how long will it take to return to the ground? Solve by completing the square. Round your answer to the nearest tenth of a second.
Mathematics
1 answer:
sesenic [268]2 years ago
3 0
At the time the rocket hits the ground h=0, given that h=-16t²+320t+32
when h=0, our equation will be:
-16t²+320t+32=0
solving the above by completing square method we proceed as follows;
-16t²+320t+32=0
divide though by -16 we get
t²-20t-2=0
t²-20t=2
but
c=(-b/2a)^2
c=(20/2)^2
c=100
hence:
t²-20t+100=100+2
(t-10)(t-10)=102
√(t-10)²=√102
t-10=√102
hence
t=10+/-√102
t~20.1 or -0.1
since it must have taken long, then the answer is 20.1 sec

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Answer:

8

Step-by-step explanation:

Given the equation 1/4 (16 + 12x) = 28, to solve this first we open the bracket using the distributive property. According to this property, given A, B and C then:

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Step 1:

1/4 (16 + 12x) = 28

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Step 2:

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Step 3:

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Answer:

For x∈ {-∞,-3} y<0, below x-axis

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x∈ {-1,4} y<0, below x-axis

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Step-by-step explanation:

f(x)=2x^{3}-26x-24

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2x^{3}-26x-24=0\\2x^{3}-2x-24x-24=0\\2x(x^{2} -1)-24(x+1)=0\\2x(x+1)(x-1)-24(x+1)=0\\(x+1)(2x^{2} -2x-24)=0\\=> x+1=0 => x_{1}=-1\\

=> 2x^{2} -2x-24=0\\=>2(x^{2} -x-12)=0\\=> x^{2} -x-12=0\\=> x^{2} -4x+3x-12=0\\=> x(x-4)+3(x-4)=0\\=> (x-4)(x+3)=0\\

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x∈ {-3,-1} y>0, above x-axis

x∈ {-1,4} y<0, below x-axis

x∈ {4,∞} y>0, above x-axis



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