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Sonja [21]
3 years ago
11

Please help me....I really need help!!

Mathematics
1 answer:
scoundrel [369]3 years ago
6 0

\dfrac{c^2-4}{6c^4+15c^3}\div\dfrac{c^2+4c+4}{12c^3+30c^2}=\dfrac{c^2-2^2}{3c^3(2c+5)}\div\dfrac{c^2+2(c)(2)+2^2}{6c^2(2c+5)}\\\\_{\text{use}\ a^2-b^2=(a-b)(a+b)\ \text{and}\ (a+b)^2=a^2+2ab+b^2}\\\\=\dfrac{(c-2)(c+2)}{3c^3(2c+5)}\cdot\dfrac{6c^2(2c+5)}{(c+2)^2}=\dfrac{(c-2)}{c}\cdot\dfrac{2}{(c+2)}=\boxed{\dfrac{2(c-2)}{c(c+2)}}

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3 years ago
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2 years ago
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sesenic [268]

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3 0
3 years ago
How to solve this elimination
aleksandrvk [35]
Firstly you have to reduce the 4 variables to a double system of equation with only 2 variables.

Let's say, first we have to eliminate Z:
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At the end you will find:
w = -3
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4 0
2 years ago
Graph f(x)=<img src="https://tex.z-dn.net/?f=%202%5E%7Bx%7D%20" id="TexFormula1" title=" 2^{x} " alt=" 2^{x} " align="absmiddle"
Lynna [10]
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