Solving Two- Step Equations:
First you have to add or subtract the number without the variable, x, then divide with the variable with x in it.
1.) -2 - 2=26
Answer: x= -14
Try some on your own, if you still need help comment back!
Answer:
n = 8.15
Step-by-step explanation:
3/8 is 4 & 3/5 % of what number?
Rewrite this as: 3/8 is (4 3/5)% of what number?
Convert 4 3/5% into a decimal: 0.046
Then write out and solve the equation 3/8 = 0.046n, where n is the unknown number.
Dividing both sides by 0.046, we get n = 8.15
1. You have that two sides of a triangle have lengths of 6 and 13. Then, by the <span>Triangle Inequality Theorem, you have:
a+b>c
a+c>b
b+c>a
3. Therefore, you have:
a=6
b=13
c<a+b
c<6+13
c<19
4. The difference between a and b should be lesser that c. Then
a-b<c
13-6<c
7<c
5. Therefore:
c=x
7<x<19
The correct answer is the option B: </span>
B. 7<x<19<span>
</span>
Answer:
940 m
Step-by-step explanation:
Above sea level is measured as positive and below sea level is measured as negative.
In order to find the distance between both we need to find the difference between their elevation and depression.
Distance between Grace and her friend
= 647 - (-293)
= 647 + 293
= 940 m
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)
