Triangles A B C and A double-prime B double-prime C double-prime are shown. Triangle A double-prime B double-prime C double-prim
1 answer:
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If this is your equation: 
Solution:
LCD for 5 and 5x is 5x
←cross product
20x + 30 = 5x² + 25x ←simplify with distributive property
0 = 5x² + 5x - 30 ←use inverse operations to collect all terms on one side
0 = x² + x - 6 ←if possible divide by numerical GCF (This case 5)
0 = (x + 3)(x - 2) ←Factor
x = -3 or x = 2
Please check by substitution in original equation... Both work
Solution:
LCD for 5 and 5x is 5x
20x + 30 = 5x² + 25x ←simplify with distributive property
0 = 5x² + 5x - 30 ←use inverse operations to collect all terms on one side
0 = x² + x - 6 ←if possible divide by numerical GCF (This case 5)
0 = (x + 3)(x - 2) ←Factor
x = -3 or x = 2
Please check by substitution in original equation... Both work
To solve this problem you must apply the proccedure shown below:
1. You have the following expression given in the problem above:
![\sqrt[3]{216 x^{27} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B216%20x%5E%7B27%7D%20%7D%20)
2. Rewriting the expression we have:
![\sqrt[3]{6^3 x^{27} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B6%5E3%20x%5E%7B27%7D%20%7D%20)
3. You have that
and the exponent 
are divisible by index 
. Therefore, you have:
![\sqrt[3]{216 x^{27} } =6 x^{9}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B216%20x%5E%7B27%7D%20%7D%20%3D6%20x%5E%7B9%7D%20)
Therefore, as you can see,
the answer is the option, which is:
1. You have the following expression given in the problem above:
2. Rewriting the expression we have:
3. You have that
Therefore, as you can see,
the answer is the option, which is: