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simplify the expressions...
1. (x^3)^5
2. (y^4)^5
3. 4^8/4^2(this is a fraction)
4. b^2/b^4(this is a fraction)
5. 28xy^7/32xy^12(this is a fraction)
Please show me step by step because I am not really understanding these types of problems...
6 answers
Wow...that was rude. I didn't mean to make it sound that way, I was just confused...
o_o sorry
o_o sorry
If you demonstrate what you've done so far and/or where you think you're running into trouble, a math tutor will be able to help you better. They need to know what you know.
And please remember that when you use all caps online, it's as if you're SHOUTING at us. Just "Math" in the subject line is all you need to put -- or better, the class you're taking, such as Algebra 2 or Trig or something specific like that.
And please remember that when you use all caps online, it's as if you're SHOUTING at us. Just "Math" in the subject line is all you need to put -- or better, the class you're taking, such as Algebra 2 or Trig or something specific like that.
Okay, my bad, so sorry. I am just having a difficult time simplifying the expressions that involve fractions w/ exponents. I have my textbook and I have been working on it for awhile but I need more guidance on how to do it...I can be a slow learner sometimes. If you can help me that would be great, but I'm guessing you are not a math tutor. In the future, I won't use ALL CAPS. Sorry again...
To be able to handle these types of exponent questions, you must have been given the basic laws of exponents, or they will be found in your text or notes.
your first question matches the rule that
(a^x)^y = a^(xy)
so , (x^3)^5 = x^15
#2 is the same
#3, you will need a^x / a^y = a^(x-y)
#4 same thing
#5, think of it as
28xy^7/(32xy^12)
= (28/32)(x/x)(y^7/y^12)
= (7/8)(y^-5)
or
= (7/8)(1/y^5) or 7/(8y^5)
using the rule that a^-n = 1/a^n
your first question matches the rule that
(a^x)^y = a^(xy)
so , (x^3)^5 = x^15
#2 is the same
#3, you will need a^x / a^y = a^(x-y)
#4 same thing
#5, think of it as
28xy^7/(32xy^12)
= (28/32)(x/x)(y^7/y^12)
= (7/8)(y^-5)
or
= (7/8)(1/y^5) or 7/(8y^5)
using the rule that a^-n = 1/a^n
Oh okay! That actually makes more sense...Thanks :)
1 answer