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KiRa [710]
3 years ago
15

I need an explanation for this PLEASEE!

Mathematics
1 answer:
SashulF [63]3 years ago
4 0

Answer:

4)Y axis :X=0

5)A:2units B:4 units C:7units D:7units

A':2units B':4units C':7units D':7units

6)They are correspondingly the same

7)A(2;1) B(4;4) C(7;4) D(7;1)

A'(-2;1) B'(-4;4) C'(-7;4) D(-7;1)

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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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2 years ago
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