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Fynjy0 [20]
3 years ago
12

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

Mathematics
2 answers:
worty [1.4K]3 years ago
7 0

Answer:

The maximum height is 1087.06 ft.

Step-by-step explanation:

We need to take the derivative of the equation and set it to 0.

y' = - 32x + 254 = 0\\32x = 254\\x = 7.9375

We now can plug x = 7.9375 into the original equation.

y=-16x^{2} +254x+79\\y=-16(7.9375)^{2}+254(7.9375)+79\\y=1087.06 ft

zimovet [89]3 years ago
6 0

Answer:

1087.1

Step-by-step explanation:

same as the other comment it's just that if your on delta math they make you round.

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sukhopar [10]
3/5 block mile is also 0.6 block mile
4 1/2 miles is also 4.5 miles

How many 0.6mi blocks do you run to run 4.5 miles?

4.5miles/0.6block mi = 7.5 or 7 1/2 blocks will equal 4.5 or 4 1/2 miles.

Check 7.5 blocks x 0.6 block mi = 4.5 miles

Same as you know one quarter equals $0.25 cent. And how many do you need to make one dollar that equals $1.00 ?

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3 years ago
What number divided by -4 =64?
notka56 [123]
The number divided by -4 to get 64 was -256
6 0
3 years ago
Please help, thank you!!
julsineya [31]

Answer:

Step-by-step explanation:

Givens

  • Triangle
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A = Base * height / 2

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4 0
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Read 2 more answers
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
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