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2 years ago

If you write 14 to the power of 6 in expanded form, how many factors will there be?

Mathematics

2 answers:

AleksandrR [38]2 years ago

7 0

9514 1404 393

Answer:

Step-by-step explanation:

The power is used to signify repeated multiplication in the same way a multiplier signifies repeated addition.

14⁶ = 14×14×14×14×14×14

There are 6 factors of 14.

Allushta [10]2 years ago

4 0

There will be 49 factors I belive!

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3 years ago

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What are two different way of factoring -3x-9? Select all that apply

Ghella [55]

Options:

A. –3(x + 3)

B. 3(x + 3)

C. 3(–x – 3)

D. –3(x – 3)

Answer:

A and C

Step-by-step explanation:

Given

$-3x - 9$

Required

Factorize

The options are not given; So, I'll answer on a general term

$-3x - 9$

Express -9 as -3 * 3

$-3x - 9 = -3x - 3 * 3$

Then factorize 3 out of the expression as:

$-3x - 9 = 3(-x - 3)$

Another way is to factor -3 out of the expression as:

$-3x - 9 = -3x - 3 * 3$

$-3x - 9 = -3(x + 3)$

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2 years ago

Find the sum of the first 3030 terms in this geometric series:

Viefleur [7K]

Answer:

I think it might be B. but I am not so sure.

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3 years ago

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Kathy starts with a number she adds 3 touring numbers then doubles the results finally she subtract 7 if she ended with nine wha

Ann [662]

she ends up getting 5

8 0

3 years ago

If z^2+1/z^2=11 , find the value of z^3-1/z^3,using only the positive value of z-1/z

vredina [299]

Hello,

Let's assume

$t=z- \dfrac{1}{z} \\\\

t^2=z^2-2+ \dfrac{1}{z^2}= 11-2=9==\ \textgreater \ t=3\\\\

t^3=z^3-3z^2* \dfrac{1}{z} +3z* \dfrac{1}{z^2} - \dfrac{1}{z^3} \\\\
=z^3 - \dfrac{1}{z^3} -3(z-\dfrac{1}{z})\\\\
z^3 - \dfrac{1}{z^3}=t^3+3t=3^3+9=27+9=36


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3 years ago