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2 years ago

Evaluate the following expression (-3)^-2

Mathematics

2 answers:

OleMash [197]2 years ago

8 0

<h3>Answer: the fraction 1/9</h3>

It is approximately equal to 0.1111

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To get this answer, we use the rule

$x^{-k} = \frac{1}{x^k}$

The negative exponent tells us to apply the reciprocal to get the exponent to be positive.

So,

$(-3)^{-2} = \frac{1}{(-3)^2}\\\\(-3)^{-2} = \frac{1}{9}$

Squaring -3 means you square the negative as well

$(-3)^2 = (-3)*(-3) = 9$

svlad2 [7]2 years ago

7 0

Answer:

1/9

Step-by-step explanation:

(-3)^-2 would be better written as (-3)^(-2).

(-3)^(-2) would be easier to evaluate if written as -------------

(-3)^2

The final answer is 1/9.

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artcher [175]

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a = (8 – 5)

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3 years ago

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3 years ago

Evaluate each finite series for the specified number of terms. 1+2+4+...;n=5

zaharov [31]

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

2*2 = 4

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.

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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: