Ray BD bisects m<ABC, solve for x and finf m<ABCm<ABD=6x-1, m<CBD=4x+7x=
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Answer:
<h3>20feet</h3>Step-by-step explanation:
Initial height of the ball = 8feet
If it bounces 60% of its previous height, its new height will be;
60% of 8
= 60/100 * 8
= 480/100
= 4.8 ft
If it bounces 60% of its current distance, new height will be expressed as;
60% of 4.8
= 0.6 * 4.8
= 2.88
The height of the ball will keep reducing and form a geometric progression of the form 8, 4.8, 2.88...
In order to get the total distance traveled by the ball, we need to calculate the sum to infinity of the sequence;
S∞ = a/1-r where;
a is the first term = 8
r is the common ratio
Substitute into the formula;
S∞ = a/1-r
S∞ = 8/1-0.6
S∞ = 8/0.4
S∞ = 20feet
Hence the total distance traveled by the ball is 20feet
Check the picture below.
so the lunchbox is really a semi-circle with a radius of 3, sitting on top of a square of 6x6, with a lenght of 10.
the volume of the lunchbox then will be the area of the semi-circle and square times the length.
![\bf \left[ \stackrel{\textit{semi-circle's area}}{\frac{\pi 3^2}{2}}+\stackrel{\textit{square's area}}{(6\cdot 6)} \right]\stackrel{length}{10}\implies \left( \frac{9\pi }{2}+36\right)(10)](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Bsemi-circle%27s%20area%7D%7D%7B%5Cfrac%7B%5Cpi%203%5E2%7D%7B2%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bsquare%27s%20area%7D%7D%7B%286%5Ccdot%206%29%7D%20%5Cright%5D%5Cstackrel%7Blength%7D%7B10%7D%5Cimplies%20%5Cleft%28%20%20%5Cfrac%7B9%5Cpi%20%7D%7B2%7D%2B36%5Cright%29%2810%29)
so the lunchbox is really a semi-circle with a radius of 3, sitting on top of a square of 6x6, with a lenght of 10.
the volume of the lunchbox then will be the area of the semi-circle and square times the length.