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Colt1911 [192]
3 years ago
15

The shape of a dome can be modeled by the equation h=2dsquared +100 when h issue height (in feet) of the dome from the floor d f

eet from its center. How far from the center of the domes is the height 50 feet?
Mathematics
1 answer:
Law Incorporation [45]3 years ago
6 0

Given that,

The shape of a dome can be modeled by the equation :

h=2d^2+100 ....(1)

h is the height of the dome

d is distance from its center

We need to find the distance from the center of the dome if the height is 50 feet. Put h = 50 in equation (1)

50=2d^2+100\\\\2d^2=100-50\\\\2d^2=50\\\\d^2=25\\\\d=5\ \text{feet}, neglecting d = -5 feet (as height can't be negative)

So, at a distance of 5 feet from the center of domes the height is 50 feet.

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<h3>Given</h3>

trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8

<h3>Find</h3>

the area of PSTK

<h3>Solution</h3>

It helps to draw a diagram.

∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as

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