1st) You would find the volume of the larger solid. V = Base Area • Height = (10•15) • 10 = 1500
2nd find the volume of the smaller figure. (2•4) • 4 = 32
Subtract the first from second 1500 - 32 = 1468.
Then you have to find the area of the fsces left after you took out figure two and add them to the total. 4•2 = 8. 4•2 = 8. 4•4 = 16
1462 + 16 + 8 + 8 = 1494 inches cubed
The answer is d.
Finding area of ABCD :
- Find side length
2. Apply formula to find area
Finding area of GHIA :
Finding area of DEFG :
Now, let's see whether is true.
- Area (ABCD) - Area (GHIA) = Area (DEFG)
- 25 - 16 = 9
- 9 = 9
∴ Hence, it is proved √
Answer:
a) Find the common ratio of this sequence.
Answer: -0.82
b) Find the sum to infinity of this sequence.
Answer: 2.2
Step-by-step explanation:
nth term in geometric series is given by ![4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{0.549} \\r = 0.82](https://tex.z-dn.net/?f=4%5C%20th%20%5C%20term%20%3D%20ar%5En-1%5C%5C-2.196%20%3D%204r%5E%7B4-1%7D%20%5C%5C-2.196%2F4%20%3D%20r%5E%7B3%7D%20%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B0.549%7D%20%5C%5Cr%20%3D%200.82)
where
a is the first term
r is the common ratio and
n is the nth term
_________________________________
given
a = 4
4th term = -2.196
let
common ratio of this sequence. be r
![4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{-0.549} \\r = -0.82](https://tex.z-dn.net/?f=4%5C%20th%20%5C%20term%20%3D%20ar%5En-1%5C%5C-2.196%20%3D%204r%5E%7B4-1%7D%20%5C%5C-2.196%2F4%20%3D%20r%5E%7B3%7D%20%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B-0.549%7D%20%5C%5Cr%20%3D%20-0.82)
a) Find the common ratio of this sequence.
answer: -0.82
sum of infinity of geometric sequence is given by = a/(1-r)
thus,
sum to infinity of this sequence = 4/(1-(-0.82) = 4/1.82 = 2.2
4,500 plus 15% is 5,175 minus 4% is 4,968 so your answer is 4,968.