Circumference= pi × d
So
d= circumference/pi
So to find out the diameter, just do 56.52/pi= 17.99mm (2 d.p.)
Answer:
volume of a cube = 441cm3.
l³=441cm³
l=7.6cm
Area is square units, so the new area would be the scale factor squared:
Scale factor is 2, so the area would increase by 2^2 = 4
The answer is B.4
Answer:
The equivalent expression to the given expression is ![\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B32x%5E8y%5E%7B10%7D%7D%3D2x%5E2y%5E3%5Csqrt%5B3%5D%7B4x%5E2y%7D)
Step-by-step explanation:
The given expression is ![\sqrt[3]{32x^8y^{10}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B32x%5E8y%5E%7B10%7D%7D)
To find the equivalent expression:
![\sqrt[3]{32x^8y^{10}}=(32x^8y^{10})^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B32x%5E8y%5E%7B10%7D%7D%3D%2832x%5E8y%5E%7B10%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
We may write the above expression as below:

(using square root properties)
(combining the like terms and doing multiplication )
Therefore ![\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B32x%5E8y%5E%7B10%7D%7D%3D2x%5E2y%5E3%5Csqrt%5B3%5D%7B4x%5E2y%7D)
Therefore the equivalent expression to the given expression is ![\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B32x%5E8y%5E%7B10%7D%7D%3D2x%5E2y%5E3%5Csqrt%5B3%5D%7B4x%5E2y%7D)