# Help with 10th grade math homework?

Topic: Help with 10th grade math homework?
May 24, 2019 / By Linden
Question: Mean Girls quote, lol. I'm stuck on these two questions, any help would be appreciated! (i) (x^6-1)/(x^2-1) (ii) (x^3-y^3)/(x^2-y^2) Oh yes, sorry. Simplify each of the following by expressing as a single fraction.

Jareth | 2 days ago
I can help! :D (Well...duh, or I wouldn't be answering your question, lol). Anyways, we learned this a few months ago, so it's still fresh! Before you start, it's easier to see if you have it written out so that the numerator is on the top instead of the side, and the denominator on the bottom if it's not already. (i) (x^6-1)/(x^2-1) Ok, so first thing. the -1 on the numerator and the -1 on the denominator cross each other out. Now you have... (x^6)/(x^2) As you probably learned, when you divide a value with exponents, you subtract the numerator exponent from the denominator exponent. Do that, and you get your answer: x^4 If you absolutely need it in a fraction, it would be (x^4)/1. (ii) (x^3-y^3)/(x^2-y^2) First, we can split this problem to get the same thing but in a different form. That looks like this: (x^3)/(x^2) (-y^3)/(-y^2) (x to the power of 3 divided by x to the power of 2, times negative y to the power of 3 divided by negative y to the power of 2). (x^3)/(x^2), when subtracting the exponents, makes x^1 or just x times (-y^3)/(-y^2). (-y^3)/(-y^2), when subtracting the exponents, makes -y^1 or just -y. Now you have x(-y). You're not finished. It's better to get rid of that negative next to the y and make it positive. To do this, find the reciprocal. ("Flipping" the value). You have your answer! (x)/(y)
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Originally Answered: Homework Help? Find Sides of Right Triangles. 10th grade math.?
1) The hypotenuse = a*sqrt(2) = 8*sqrt(10)*sqrt(2) = 35.78. You did it right. 2) The side opposite to 30 is half of 30, which is 15. You did it right. 3) Tthe hypotenuse = a*sqrt(2) = 1.5*sqrt(2) = 2.12. Double check your work.

Freddy
Assuming that you just want to do the divisions, for the first one you could just try the long-division algorithm : . . . . . . . . . x^4 . . . + . . . x^2 . . . . + . . . .1 . . . . . . . . . . . . . . . __________________________________ x² + 0x - 1 ) x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 1 . . . . . . . . . x^6 + . 0 . .- x^4 . . . . . . . . . ______________ . . . . . . . . . . . . . . . . . . . x^4 + 0x^3 + 0x^2 . . . . . . . . . . . . . . . . . . . x^4 + . 0 . - . x^2 . . . . . . . . . . . . . . . . . . .______________ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x^2 + 0x - 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x^2 +. 0 - 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _________ So (x^6 - 1) / (x^2 - 1) = x^4 + x^2 + 1, which you should check by multiplying out (x^4 + x^2 + 1) (x^2 - 1). The second one you should recognise as "the difference of two cubes" / "the difference of two squares", which can be factorised (memorise these !) x^3 - y^3 = (x - y) (x^2 + xy + y^2) x^2 - y^2 = (x - y) (x + y) And you can see that there is a common factor of (x - y) which you can "cancel", leaving just (x^2 + xy + y^2) / (x + y) Which is in its lowest terms : you cannot reduce it any further.
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Delaia
Well basically you're going to use some factoring especially difference of cubes [*see source]. Recall the formula: (x^3 - y^3) = (x - y) * (x^2 + xy + y^2) (i) (x^6 - 1) / (x^2 - 1) Notice how we can rewrite the above as shown below: [ (x^2)^3 - 1^3 ] / (x^2 - 1) [ (x^2 - 1) * (x^4 + x^2 + 1) ] / (x^2 - 1) .: x^4 + x^2 + 1 (ii) (x^3 - y^3) / (x^2 - y^2) Besides the difference of cubes, you can also apply difference of squares. [ (x - y) * (x^2 + xy + y^2) ] / [ (x + y) * (x - y)] .: (x^2 + xy + y^2) / (x + y)
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Barzilai
care to give any directions on what is it to be done? I =0? Simplify? Find all roots? what are the shared solutions of the two? etc?
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