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Galina-37 [17]
3 years ago
6

A projectile is fired upward from the ground with an initial velocity of 300 feet per second. Neglecting air resistance, the hei

ght of the projectile at any time t can be described by the polynomial function P(t) = ­16t 2 + 300t
Mathematics
1 answer:
levacccp [35]3 years ago
8 0

Answer: The polynomial that describes the projectile's height is:

P(t) = - ­16t*2 + 300t

For t = 1 second we have:

P(1) = - 16*1^2 + 300*1

P(1) = 284 ft

The height of the projectile after 1 second is 284 feet.

Step-by-step explanation:

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Step-by-step explanation:

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andrew-mc [135]

Answer:

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•°• LHS = RHS

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The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

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x = -1

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To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

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\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

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Answer:

the period of this graph is 2\pi

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One is at x = 0 and the next is at x = 2\pi

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