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3 years ago

Convert the following repeating decimal to fractions 0.123232323...

1 answer:

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:

Neko [114]

128.309728

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

find the difference of each number and 67.92(the mean)

add all these numbers together and divide by 50

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3 years ago

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

Step-by-step explanation:

total price = $ 2.5 + $3.4(6)= $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 $\frac{10}{10}$

$0.6*\frac{10}{10} =\frac{6}{10}$

step 2

Divide numerator and denominator by $2$

$\frac{\frac{6}{2}}{\frac{10}{2}} =\frac{3}{5}$

therefore

the answer is

$\frac{3}{5}$

4 0

3 years ago

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Examine the power.

son4ous [18]

La respuesta es el número 4

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.

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3 years ago