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LenKa [72]
3 years ago
9

Noah has a coupon for 30% off at his favorite clothing store he utilizes it to buy a hoodie and a pair of jeans

Mathematics
1 answer:
Genrish500 [490]3 years ago
8 0

Answer:

$42.00 28.00 times 30% equals 42.00

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Y= -4x+9; (-5, ), (2, ), (4, )
Zolol [24]

Answer:

(-5,29)

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4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%
Naddik [55]

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)

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}}

6 0
1 year ago
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