Czarcasm

\((\frac{11}{5}, \frac{33}{5})\)

I'm assuming no one's helped you yet because the question is too confusing.

A) Is it \(\sqrt[3]{8x} \cdot \sqrt{16x^2}\)

B) Or is it \(\sqrt{8x} \cdot \sqrt{16x^2}\)

C) Or is it \(\sqrt[3]{8x} \cdot \sqrt[3]{16x^2}\)

D) Or is it ​\(\sqrt{8x} \cdot \sqrt[3]{16x^2}\)

E) Or is it neither?

A = \(\sqrt[3]{8} \cdot \sqrt[3]{x} \cdot \sqrt{16} \cdot \sqrt{x} = 2\sqrt[3]{x} \cdot 4x = \boxed{8x\sqrt[3]{x}}\)

B = \(\sqrt{4} \cdot \sqrt{2x} \cdot \sqrt{16} \cdot \sqrt{x^2} = \boxed{8x\sqrt{2x}}\)

C = ​​\(\sqrt[3]{8} \cdot \sqrt[3]{x} \cdot \sqrt[3]{8} \cdot \sqrt[3]{2x^2} = \boxed{4x\sqrt[3]{3}}\)

D = \(\boxed{4\sqrt{2x} \cdot \sqrt[3]{2x^2}}\) 

E = Well, I can't help you there, but next time make your question clearer

You have to imagine the + as +1. So it would be:

\(1(-4) = 1 \times (-4) = -4\)

Because of the Identity Property of Multiplication, anything multiplied by 1 is itself, so it follows for -4 as well.

That means our answer is: \(\boxed{-4}\)

It follows that for any right triangle, you can find the length of the hypotenuse using Pythagorean Theorem. Basically, you add the squares of the legs to get the square of the hypotenuse. If one leg was A, the other was B, and the hypotenuse was C, ​\(A^2 + B^2 = C^2\)

So if A was 3, B was 4, then C would be the square root of 9 + 16 = 25, so the hypotenuse would be 5.

Here's a formula for C, the hypotenuse.

\(C = \sqrt{A^2 + B^2}\)

Find more about the Pythagorean Theorem here: