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Amanda [17]
3 years ago
11

The height in feet of a rocket launched into the air is modeled by the function (t)=-16t^2+160t+300 where t is time in seconds.

Approximately how many seconds will the rocket exceed the height of 500 ft?
Mathematics
1 answer:
liq [111]3 years ago
7 0

approximately 1.5 seconds

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Step-by-step explanation:

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From this formula, we can derive a specific one that will serve for any value of <em>t</em>.

a)

A=5000{(1+0.03)}^{t}\\A=5000{(1.03)}^{t}

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Which algebraic expression has like terms?
iris [78.8K]

Answer:

Answer:

safe speed for the larger radius track u= √2 v

Explanation:

The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.

Also given that r_1= smaller radius

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r_2= 2r_1..............i

let u be the speed of larger radius curve

now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=

r

1

mv

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=

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mu

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form i and ii we can write

v^2= \frac{1}{2} u^2v

2

=

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