Answer:
![\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }](https://tex.z-dn.net/?f=%5Cboxed%7B5%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%7D)
Step-by-step explanation:
![\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)
![\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%20%5Cimplies%20%2810%5E%5Cfrac%7B1%7D%7B3%7D%20%29%5E%5Cfrac%7B1%7D%7B2%7D%20%3D10%5E%5Cfrac%7B1%7D%7B6%7D%20%3D%5Csqrt%5B6%5D%7B10%7D)
![\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%3D%5Csqrt%5B6%5D%7B10%7D)
![\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Ctext%7BSolving%20%7D%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)

![\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%3D%5Csqrt%5B3%5D%7B2%5Ccdot%205%5E3%7D%3D5%20%20%5Csqrt%5B3%5D%7B2%7D)
Once
![\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%20%5Csqrt%5B6%5D%7B10%7D)
We have
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
We can proceed considering the common base of exponentials
![\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%3D%20%202%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Ccdot%20%202%5E%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%20%20%3D%202%5E%7B%5Cfrac%7B3%7D%7B6%7D%20%7D%20%3D%202%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%3D%5Csqrt%7B2%7D)
Therefore,
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%205%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
5 is the y intercept.
-Always the number without the variable.
Answer:
1000 cm^3
Step-by-step explanation:
First, Find the side length of the cube. The surface area is all the area on the surface of a 3-dimensional object. Since a square has 6 congruent faces the formula for surface area is 6s^2 where s is the side length. Since we know the surface area we can solve for the side length.
600=6s^2- DIvide by 6
100=s^2- Take the square root of both sides.
10=s
The volume of a cube is s^3.
10^3=V
1000=V
I think your answer is A. (0,5)
Hope this helps!! Have a great day!! :)))
Answer:
A. 13%
Step-by-step explanation:
let's suppose there's 1000 bars of chocolate in general
70 % dark = 700 dark choco bars
30 % milk = 300 milk choco bars
20 % have nuts = 200 bars have nuts
80 % don't have nuts = 800 bars don't have nuts
if 90 % of dark chocolate bars don't have nuts, that means 630 dark chocolate bars don't have nuts, but 70 have nuts
200 bars in general have nuts, if only 70 dark chocolate bars have nuts,
200 - 70 = 130
130 = 13%