<h3>
Answer:</h3>
b. The side length of a cube is dependent on the volume.
<h3>
Step-by-step explanation:</h3>
The function definition V(s) means the volume (V) is dependent upon the side length (s), <em>not the other way around</em>.
Answer:
8x + 20 ≥ 25
Step-by-step explanation:
Well 8 times the sum of a number "x" plus 20,
So we can write the following,
8x + 20
and that should be greater than or equal to 25
8x + 20 ≥ 25
<em>Thus,</em>
<em>as an inequality it is 8x + 20 ≥ 25.</em>
<em />
<em>Hope this helps :)</em>
Answer:
30
Step-by-step explanation:
is going to output a number in between 
and 
.
Luckily you are looking for 
in the 1st quadrant which is contained in the given interval above, so the answer just comes down to what the calculator returns from entering 
in your calculator.
You may also use the unit circle. sine value refers to the y-coordinate on there.
So you are looking for when the y-coordinate is 1/2 in the fist quadrant which is at 30 degrees.
Answer:
2x + 7y + 37
Step-by-step explanation:
9+2(x+7)+7(y+2)
= 9 + 2x + 14 + 7y + 14
= 2x + 7y + 37
We have
![\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)](https://tex.z-dn.net/?f=%5Csqrt%5Bk%5D%7B%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%7D%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%5Cright%29%7Dk%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%5Cright%29%2B%5Cln%5Cleft%28%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%5Cright%29%2B%20%5Ccdots%20%2B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%5Cright%29%7Dk%5Cright%29)
and as k goes to ∞, the exponent converges to a definite integral. So the limit is
![\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bk%5Cto%5Cinfty%7D%20%5Csqrt%5Bk%5D%7B%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%7D%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Clim_%7Bk%5Cto%5Cinfty%7D%20%5Cfrac1k%20%5Csum_%7Bi%3D1%7D%5Ek%20%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cfrac%20ik%5Cright%29%5Cright%29%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cint_0%5E1%20%5Cln%5Cleft%28%5CGamma%28x%29%5Cright%29%5C%2C%20dx%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%282%5Cpi%29%7D2%7D%5Cright%29%20%3D%20%5Cboxed%7B%5Csqrt%7B2%5Cpi%7D%7D)
