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3 years ago

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Mathematics

2 answers:

Alenkinab [10]3 years ago

8 0

Answer:

Step-by-step explanation:

To solve this, you can add 2/5 b to both sides of the equation. You now have 3 + 5/5 (or 1) b = 11. => 3 + b = 11 => Now, subtract 3 from both sides, and you have b = 11 - 3, and then b = 8.

taurus [48]3 years ago

6 0

Step-by-step explanation:

3 + 2/5 b =11 - 2/5 b (this is the given question is it?)

2/5 b + 2/5 b= 11 - 3 (like terms together, you transpose -2/5b to the other side as shown.)

2b+2b = 8 ( at that stage you would find t

5 hat 5 is the common value to go into 5 itself, as shown.

4b = 8 ( you just add 2b + 2b to have 4b)

4b = 8 ( at that point you cross multiply

5 1 as shown)

4b = 8 × 5 ( simple math as shown)

4b = 40 ( you multiply 8 × 5 to obtain 40)

b =10 you can prove that. Thank you.

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Arrange the geometric series from least to greatest based on the value of their sums.

son4ous [18]

Answer:

80 < 93 < 121 < 127

Step-by-step explanation:

For a geometric series,

$\sum_{t=1}^{n}a(r)^{t-1}$

Formula to be used,

Sum of t terms of a geometric series = $\frac{a(r^t-1)}{r-1}$

Here t = number of terms

a = first term

r = common ratio

1). $\sum_{t=1}^{5}3(2)^{t-1}$

First term of this series 'a' = 3

Common ratio 'r' = 2

Number of terms 't' = 5

Therefore, sum of 5 terms of the series = $\frac{3(2^5-1)}{(2-1)}$

= 93

2). $\sum_{t=1}^{7}(2)^{t-1}$

First term 'a' = 1

Common ratio 'r' = 2

Number of terms 't' = 7

Sum of 7 terms of this series = $\frac{1(2^7-1)}{(2-1)}$

= 127

3). $\sum_{t=1}^{5}(3)^{t-1}$

First term 'a' = 1

Common ratio 'r' = 3

Number of terms 't' = 5

Therefore, sum of 5 terms = $\frac{1(3^5-1)}{3-1}$

= 121

4). $\sum_{t=1}^{4}2(3)^{t-1}$

First term 'a' = 2

Common ratio 'r' = 3

Number of terms 't' = 4

Therefore, sum of 4 terms of the series = $\frac{2(3^4-1)}{3-1}$

= 80

80 < 93 < 121 < 127 will be the answer.

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soldier1979 [14.2K]

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Answer:

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Step-by-step explanation:

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