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IrinaVladis [17]
3 years ago
12

Convert the following repeating decimal to fractions 0.123232323...

Mathematics
1 answer:
liq [111]3 years ago
6 0

Answer by Mimiwhatsup: Convert the decimal numbers to a fraction by placing the decimal numbers over a power of ten. Since there are 13 numbers to the right of the decimal point, place the decimal numbers over 10^13 (10000000000000).

\frac{1232323232323}{10000000000000}




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What is the Mean Absolute Deviation:
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128.309728

add all numbers together and divide by 50(there are 50 numbers)

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In the past month, Josh rented one video game and six DVDs. The rental price for the video game who is $2.50. The rental price f
dolphi86 [110]

Answer:

total amount paid = $ 22.9

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What is 0.6 as a fraction in simplest form
Illusion [34]

we know that


step 1

multiply 0.6 by \frac{10}{10}

so

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step 2

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\frac{3}{5}

4 0
3 years ago
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Examine the power.
son4ous [18]
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6 0
3 years ago
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Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
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