well, let's say 1 meter is the 100%, now, if we know that 1 is 100%, what percentage is 70?

Answer:
A is the answer
Step-by-step explanation:
Anytime a number is being divided by 10 it moves to the left creating a decimal.Example : 5÷10 = 0.5 , 500÷10 = 50
Anytime a number is being multiplied by 10 the number increases with a 0. Example : 5 × 10 =50
I hope this helped
Answer:
try to work this out your self then use quiz let if you can
Step-by-step explanation:
Taking

and differentiating both sides with respect to

yields
![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have

Differentiating again gives
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have

Now, when

and

, we have
Answer: 
Step-by-step explanation:
Based on the picture I assume that the exercise asks for the diameter of the sphere.
So, it is necessary to remember the following:
1. the volume of a sphere can be calculated with the following formula:

Where "r" is the radius.
2. The radius is twice the diameter.
So, knowing that the volume of the sphere is:

You can substitute them into the equation and then solve for "r" in order to find its value:
![9,145.33\pi\ in^3=\frac{4}{3}\pi r^3\\\\\frac{(3)(9,145.33\pi\ in^3)}{4\pi}=r^3\\\\\sqrt[3]{\frac{(3)(9,145.33\pi\ in^3)}{4\pi}}=r\\\\r=19\ in\\\\](https://tex.z-dn.net/?f=9%2C145.33%5Cpi%5C%20in%5E3%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%5C%5C%5C%5C%5Cfrac%7B%283%29%289%2C145.33%5Cpi%5C%20in%5E3%29%7D%7B4%5Cpi%7D%3Dr%5E3%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B%283%29%289%2C145.33%5Cpi%5C%20in%5E3%29%7D%7B4%5Cpi%7D%7D%3Dr%5C%5C%5C%5Cr%3D19%5C%20in%5C%5C%5C%5C)
Then, the diameter is:
