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M1n3Work
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I was wondering if anyone could possibly help me with this math problem, I had passed algebra long ago, but I had this problem in my other memory test, which I had forgotten..
c^9 d^7/c^14 d^10
a > d^2/d^4
b > c5d3
c > d^3/c^5
d > c2d4
I'm not really sure how this website works, but credit or reward will be the one that is helpful towards me. Thanks.
 one year ago
 one year ago
M1n3Work Group Title
I was wondering if anyone could possibly help me with this math problem, I had passed algebra long ago, but I had this problem in my other memory test, which I had forgotten.. c^9 d^7/c^14 d^10 a > d^2/d^4 b > c5d3 c > d^3/c^5 d > c2d4 I'm not really sure how this website works, but credit or reward will be the one that is helpful towards me. Thanks.
 one year ago
 one year ago

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AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
\[ \large \frac{c^9 d^{7}}{c^{14} d^{10}} \] There are two properties we can use here: 1) Negative Exponents: \(\large a^{x} = \frac{1}{a^{x}} \) We may take a negative exponent over or under the fraction bar to get rid of the negative / turn it into a positive. 2) Multiplication of likebases: \(\large a^{x} a^{y} = a^{x + y} \) We can group together multiplication of likebases with this property, just adding together the exponents. Division is also possible if we use the first property with this one: \(\large \frac{a^{x}}{a^{y}} = a^{x}a^{y} = a^{xy}\)
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
So there are a number of ways to do this problem using some combination of those two properties. One that might be easier visually is to take everything to the numerator using the negative exponent property, add together the exponents of the likebases, and then put negative exponents back into the denominator.
 one year ago

M1n3Work Group TitleBest ResponseYou've already chosen the best response.0
Oh.. ! I see what you did there. Thanks !
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
You're welcome. :)
 one year ago
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