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katrin2010 [14]
3 years ago
13

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+218x+75
Mathematics
1 answer:
stealth61 [152]3 years ago
8 0

Answer:

6.81s

Step-by-step explanation:

First you must know that the velocity of the body is zero at the maximum height.

v =dy/dx

v= -32x+218

Since v = 0

-32x+218 =0

32x =218

x= 218/32

x = 6.81s

Hence the rocket reaches its maximum height after 6.81s

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50.27 units²

Step-by-step explanation:²

The standard equation of a circle with center at the origin is x² + y² = r², where r is the radius.  Substituting 4 for x and 0 for y yields 4² + 0² = r², so we see immediately that r = 4 units.

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Here, with r = 4, the area is A = π(4 units)² = 16π units², or

50.27 units² to the nearest hundredth.

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AB is the angle bisector of ∠CAD. Solve for x.
Sliva [168]

Answer: x=7

Step-by-step explanation:

Since AB is a bisector, it cuts the angle in half, and since we know that the angle is a right angle, we know it’s 90 degrees, so knowing that you use the equation 7x-4= 45 (since 45 is half of 90), add 4 to both sides so you have 7x=49, the divide 49 by 7 and you get 7, so x= 7

3 0
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A right circular cone is undergoing a transformation in such a way that the radius of the cone is increasing at a rate of 1/2 in
Ivenika [448]

Answer:

The volume is decreasing at the rate of 1.396 cubic inches per minute

Step-by-step explanation:

Given

Shape: Cone

\frac{dr}{dt} =\frac{1}{2} --- rate of the radius

\frac{dh}{dt} =-\frac{1}{3} --- rate of the height

r = 2

h = \frac{1}{3}

Required

Determine the rate of change of the cone volume

The volume of a cone is:

V = \frac{\pi}{3}r^2h

Differentiate with respect to time (t)

\frac{dV}{dt} = \frac{\pi}{3}(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})

Substitute values for the known variables

\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)

\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}

\frac{dV}{dt} = -\frac{22}{7*3}*\frac{4}{3}

\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}

\frac{dV}{dt} = -\frac{88}{63}

\frac{dV}{dt} =-1.396in^3/min

The volume is decreasing at the rate of 1.396 cubic inches per minute

3 0
3 years ago
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