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Mathematics 13 Online
OpenStudy (anonymous):

Help! Could somebody please explain!? I will give a medal!(: 1. Which expression is equivalent to 6^-3 A. 1/6^3 B. -6^3 C. 1/3^6 D. 6/1^6 I guess I'm just lost on 1. How to find the answer. 2. How to do negative exponents If somebody gave me the answer and then explained how they got it, i think it would help a lot. thank you <3

OpenStudy (anonymous):

A

OpenStudy (anonymous):

Okay how did you get that ?

OpenStudy (jacobbenvenutty):

A negative exponent is how many times to divide by that number. Basically all you have to do is flip the base to the other side. For instance, "x–2" (ecks to the minus two) just means "x2, but underneath, as in 1/(x2)".

OpenStudy (jacobbenvenutty):

"x2"

OpenStudy (anonymous):

6^-3= 1/216 1/6^3= 1/216

OpenStudy (anonymous):

Answer is A

OpenStudy (jacobbenvenutty):

Yep ; )

OpenStudy (anonymous):

So you would divide 6 by -3 ?

OpenStudy (jacobbenvenutty):

http://www.mathsisfun.com/algebra/negative-exponents.html

OpenStudy (jacobbenvenutty):

example: 5(-3 power) = 1 ÷ 5 ÷ 5 ÷ 5 = 0.008

OpenStudy (jacobbenvenutty):

That make any sense Mc?

OpenStudy (anonymous):

Ohhh okay i think i get it . So like 6^-3 i would be 1 divide by 6 divide by 6 divide by 6 ?

OpenStudy (jacobbenvenutty):

Yes!

OpenStudy (anonymous):

and when i did that i got 0.00462963..? is that right ?

OpenStudy (jacobbenvenutty):

An easier way to calculate it is if you /1 Example: 1 ÷ (5 × 5 × 5) = 1/53 = 1/125 = 0.008

OpenStudy (jacobbenvenutty):

Just look at the website I posted and it will explain it there

OpenStudy (anonymous):

I am looking at the website lol.

OpenStudy (anonymous):

i guess im having a problem with finding the " 1/53" in the equation 1 / ( 5 x 5 x 5)

OpenStudy (jacobbenvenutty):

So 1÷ (6 × 6 × 6)= 1/216 Going by that A is the correct answer

OpenStudy (anonymous):

Oh ohkay!

OpenStudy (jacobbenvenutty):

Make sense now?

OpenStudy (anonymous):

Yes! THANK YOU<3 Can anybody help me with "Simplifying expressions" i think i know how but, i need a little bit of guidance . i will open a new questions

OpenStudy (anonymous):

@Jacobbenvenutty

OpenStudy (jacobbenvenutty):

Sure i'll try my best ; )

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