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3 years ago

PLEASE HELP SOLVE THIS!!!!!!!!!!!

Mathematics

1 answer:

Sphinxa [80]3 years ago

4 0

9514 1404 393

Answer:

a) yes; 12/15/17 ~ 20/25/x; SAS

b) x = 28 1/3

Step-by-step explanation:

The left-side segments are in the ratio ...

top : bottom = 12 : 8 = 3 : 2

The right side segments are in the ratio ...

top : bottom = 15 : 10 = 3 : 2

These are the same ratio, and the angle at the peak is the same in both triangles, so the triangles are similar by the SAS postulate.

Normally, a similarity statement would identify the triangles by the labels on their vertices. Here, there are no such labels, so we choose to write the statement in terms of the side lengths, shortest to longest:

12/15/17 ~ 20/25/x

The sides of similar triangles are proportional, so the ratio of longest to shortest sides will be the same in the two triangles. In the smaller triangle, the longest side is 17/12 times the length of the shortest side. The value of x will be 17/12 times the length of the shortest side in the larger triangle:

x = 17/12 · 20 = 340/12

x = 28 1/3

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shows correct substitution of the values a, b, and c from the given quadratic equation into quadratic formula.

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Answer:

Step-by-step explanation:

The series are given as geometric series because these terms have common ratio and not common difference.

Our common ratio is 2 because:

1*2 = 2

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The summation formula for geometric series (r ≠ 1) is:

$\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}}$ or $\displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}$

You may use either one of these formulas but I’ll use the first formula.

We are also given that n = 5, meaning we are adding up 5 terms in the series, substitute n = 5 in along with r = 2 and first term = 1.

$\displaystyle \large{S_5=\frac{1(2^5-1)}{2-1}}\\\displaystyle \large{S_5=\frac{2^5-1}{1}}\\\displaystyle \large{S_5=2^5-1}\\\displaystyle \large{S_5=32-1}\\\displaystyle \large{S_5=31}$

Therefore, the solution is 31.

__________________________________________________________

Summary

If the sequence has common ratio then the sequence or series is classified as geometric sequence/series.

Common Ratio can be found by either multiplying terms with common ratio to get the exact next sequence which can be expressed as $\displaystyle \large{a_{n-1} \cdot r = a_n}$ meaning “previous term times ratio = next term” or you can also get the next term to divide with previous term which can be expressed as: