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exis [7]
3 years ago
15

Riley has 3and 1/4 bag of soil to put in her garden. Each bag of soil will cover 125.3 ft3 . How many cubic feet will Riley be a

ble to cover if she uses all of these bags of soil.
Mathematics
2 answers:
pogonyaev3 years ago
8 0

To solve this problem you must apply the proccedure shown below:

1. You have that Riley has:

3^{\frac{1}{4} }=3.25 bags of soil

2. Each bag of soil will covers 125.3 ft³. Therefore, if she uses all the bags she will be able to cover:

(3.25)(125.3ft^{3})=407.22ft^{3}

Therefore, the answer is: 407.22 ft³.

aliina [53]3 years ago
4 0

Hello!

Riley has 3and 1/4 bag of soil to put in her garden. Each bag of soil will cover 125.3 ft3 . How many cubic feet will Riley be able to cover if she uses all of these bags of soil.

3 1/4 = 3.25

125.3 x 3.25 = 407.225 ft.^3

Answer - 407.225 ft.^3

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Which of the following is a geometric series? 40 + 50 + 60 + 70 40 + 42 + 44 + 46 40 + 80 + 40 + 120 40 + 20 + 10 + 5
Aloiza [94]

Answer:

40 + 20 + 10 + 5 is a geometric series

Step-by-step explanation:

* Lets revise the geometric series

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- Ex:

# 5  ,  10  ,  20  ,  40  ,  80  ,  ………………………. (×2)

# 5000  ,  1000  ,  200  ,  40  ,  …………………………(÷5)

* General term (nth term) of a Geometric Series:

∵ U1 = a  ,  U2  = ar  ,  U3  = ar2  ,  U4 = ar3  ,  U5 = ar4

∴ Un = ar^n-1, where a is the first term , r is the constant ratio

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* Now lets check the answers to find the correct one

# 40 + 50 + 60 + 70

∵ 50/40 = 5/4

∵ 60/50 = 6/5

∵ 5/4 ≠ 6/5

- No common ratio between each two consecutive terms

∴ Not geometric series

# 40 + 42 + 44 + 46

∵ 42/40 = 21/20

∵ 44/42 = 22/21

∵ 21/20 ≠ 22/21

- No common ratio between each two consecutive terms

∴ Not geometric series

# 40 + 80 + 40 + 120

∵ 80/40 = 2

∵ 40/80 = 1/2

∵ 2 ≠ 1/2

- No common ratio between each two consecutive terms

∴ Not geometric series

# 40 + 20 + 10 + 5

∵ 20/40 = 1/2

∵ 10/20 = 1/2

∵ 5/10 = 1/2

- There is a common ratio between each two consecutive terms

∴ 40 + 20 + 10 + 5 is a geometric series

7 0
3 years ago
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