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mafiozo [28]
3 years ago
13

An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equati

on of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 64t + 80?
Mathematics
1 answer:
Free_Kalibri [48]3 years ago
4 0
Max height is the vertex
convert to vertex form (y=a(x-h)^2+k) by completeing the square

h(t)=-16(t²-4t)+80
h(t)=-16(t²-4t+4-4)+80
h(t)=-16((t-2)²-4)+80
h(t)=-16(t-2)²+64+80
h(t)=-16(t-2)²+144

vertex is (2,144)
at t=2, the height is 144

max height is 144ft
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3 years ago
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Maurinko [17]

Answer:

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3 years ago
X^2+y^2+14x+10y=7 <br> center and radius
Sholpan [36]
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x^2+y^2+14x+10y=7\\\\x^2+14x+y^2+10y=7\\\\x^2+2x\cdot7+y^2+2y\cdot5=7\\\\\underbrace{x^2+2x\cdot7+7^2}_{Use\ (*)}-7^2+\underbrace{y^2+2y\cdot5+5^2}_{Use\ (*)}-5^2=7\\\\(x+7)^2-49+(y+5)^2-25=7\\\\(x+7)^2+(y+5)^2-74=7\ \ \ \ |add\ 74\ to\both\ sides\\\\(x+7)^2+(y+5)^2=81\\\\(x+7)^2+(y+5)^2=9^2

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3 years ago
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<h3>Answer: Bottom right corner (ie southeast corner)</h3>

This 3D solid is a strange sideways bowl shape. Each cross section is a ring to show the empty space.

======================================================

Explanation:

Check out the diagram below. The graph was created with GeoGebra. We have y = x^2 in red and x = y^2 in blue.

The gray region is the region between the two curves. We spin this gray region around the horizontal green line y = 1 to generate the answer mentioned above.

Note how (1,1) is a fixed point that does not move as this is on the line y = 1. Every other point moves to sweep through 3D space to create the solid figure. One way you can think of it is to think of propeller blades. Or you can think of a revolving door (the door is "flat" so to speak, but it sweeps out a 3D solid cylinder).

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