Answer: If you need to find the length of the hypotenuse of a right triangle, you can use the Pythagorean theorem if you know the length of the other two sides. Square the length of the 2 sides, called a and b, then add them together. Take the square root of the result to get the hypotenuse
Question 1: To find volume the formula is Length× Width × height
So it would be 2 1/2× 6 × 3 1/2= 5 1/2
So the volume is 5 1/2
Question 2: You have to multiply 6 × 4 = 24
![2x^3+5y^3=\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right)](https://tex.z-dn.net/?f=2x%5E3%2B5y%5E3%3D%5Cleft%28%5Csqrt%5B3%5D%7B2%7Dx%2B%5Csqrt%5B3%5D%7B5%7Dy%5Cright%29%5Cleft%28%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Cright%29%5E2x%5E2-%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7B5%7Dxy%2B%5Cleft%28%5Csqrt%5B3%5D%7B5%7D%5Cright%29%5E2y%5E2%5Cright%29)
![\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right) \ \textgreater \ Refine](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%5B3%5D%7B2%7Dx%2B%5Csqrt%5B3%5D%7B5%7Dy%5Cright%29%5Cleft%28%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Cright%29%5E2x%5E2-%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7B5%7Dxy%2B%5Cleft%28%5Csqrt%5B3%5D%7B5%7D%5Cright%29%5E2y%5E2%5Cright%29%20%5C%20%5Ctextgreater%20%5C%20%20Refine)
![\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(2^{\frac{2}{3}}x^2-\sqrt[3]{10}xy+5^{\frac{2}{3}}y^2\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%5B3%5D%7B2%7Dx%2B%5Csqrt%5B3%5D%7B5%7Dy%5Cright%29%5Cleft%282%5E%7B%5Cfrac%7B2%7D%7B3%7D%7Dx%5E2-%5Csqrt%5B3%5D%7B10%7Dxy%2B5%5E%7B%5Cfrac%7B2%7D%7B3%7D%7Dy%5E2%5Cright%29)
