Answer:
Step-by-step explanation:
We'll take this step by step. The equation is
![8-3\sqrt[5]{x^3}=-7](https://tex.z-dn.net/?f=8-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-7)
Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:
![-3\sqrt[5]{x^3}=-15](https://tex.z-dn.net/?f=-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-15)
The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:
![\sqrt[5]{x^3}=5](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E3%7D%3D5)
Let's rewrite that radical into exponential form:
If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:
On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:
![x=\sqrt[3]{5^5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E5%7D)
Let's group that radicad into groups of 3's now to make the simplifying easier:
![x=\sqrt[3]{5^3*5^2}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E3%2A5%5E2%7D)
because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:
![x=5\sqrt[3]{5^2}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B5%5E2%7D)
which is the same as:
![x=5\sqrt[3]{25}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B25%7D)
C is false. 9 multiplied by 7 equals 63.
For the square root of a negative number, you would have to get into imaginary numbers. If you haven't learned about it yet, then the answer would be no real solutions. But if you have, here's how to solve it:
First, identify that 9 x 3 = 27. To after taking the square root, you will get 3√3 (Since the square root of 9 is 3 and there's no square root of 3, so leave it in the √)
However, since it is a negative number, you will have to include the letter i in the answer. i represents the imaginary part of it since you can't have a negative number in the square root.
So your answer will look like 3i√3.
Let 
be the volume of 20% solution, 
the volume of the 60% solution. We want a total volume of 400 mL in the final mixture, so
Each mL of either solution will contribute a corresponding concentration, and in the final mixture we want the 400 mL to have a 40% concentration, which means we should also have
Solve the system and we get 
.
