Answer:well If it says its wrong then it may be wrong some times is you think its right its not always right
Answer:
A
Step-by-step explanation:
So we have the equation:
First, let's subtract 16 from both sides:
Now, let's divide both sides by 3:
Remember that with fractional exponents, we can move the denominator into the root position. Therefore:
![(\sqrt[3]{x-4})^4=16](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bx-4%7D%29%5E4%3D16)
Let's take the fourth root of both sides. Since we're taking an even root, make sure to have the plus-minus symbol!
![\sqrt[3]{x-4} =\pm 2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx-4%7D%20%3D%5Cpm%202)
Cube both sides. Since we're cubing, the plus-minus stays.
Add 4 to both sides.
Calculator:
So, our answer is A.
And we're done!
564/24 equals to 47/2 or 23.5
Start with 180.
<span>Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90. </span>
<span>Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45. </span>
<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>
<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>
<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
<span>Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1 </span>
<span>Once you end up with a quotient of "1" you're done. </span>
<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>
