From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2
The answer is 91 because the tens place has to be 8 more so in the 80's, 8 would be 8 more than 0 and 0 is not one of the digits and it's a 2 digit number so it can't go to the 100's so 91 is the best answer. I hope this helped you
Answer:
c
Step-by-step explanation:
I don't trust myself so I am sorry if it is wrong
Step-by-step explanation:
6x2 - x + 3 - (-5x2 + 10x - 1)
6x2 - x + 3 + 5x2 - 10x + 1
Solving like terms
11x2 - 11x + 4
Here is one way to solve for x.
Step 1) 2x^2-7=9
Step 2) 2x^2-7+7=9+7
Step 3) 2x^2=16
Step 4) (2x^2)/2=16/2
Step 5) x^2=8
Step 6) sqrt(x^2)=sqrt(8)
Step 7) |x|=sqrt(8)
Step 8) x=sqrt(8) or x=-sqrt(8)
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Below are explanations/reasons to each of the steps above.
Step 1) Original equation
Step 2) Add 7 to both sides
Step 3) Combine like terms
Step 4) Divide both sides by 2
Step 5) Simplify
Step 6) Apply the square root to both sides. The notation "sqrt" is shorthand for "square root"
Step 7) Use the rule that sqrt(x^2) = |x| for all real numbers x
Step 8) Use the rule that if |x| = k then x = k or x = -k for some fixed number k.
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The two solutions are
x = sqrt(8) or x = -sqrt(8)
