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3 years ago

5 76/10 x 1.75 i need help

Mathematics

2 answers:

alekssr [168]3 years ago

7 0

Answer:

22.05

Step-by-step explanation:

kirill115 [55]3 years ago

7 0

Answer:

22.05

Step-by-step explanation:

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Which expressions are equivalent to 2(b+3c)?

Rina8888 [55]

Answer:

B. (b+3c)+(b+3c)

C. 2(b)+2(3c)

Step-by-step explanation:

we have

$2(b+3c)$

Distribute the number 2

$2(b+3c)=2b+2(3c)=2b+6c$

Verify each case

case A) 3(b+2c)

distribute the number 3

$3(b+2c)=3b+3(2c)=3b+6c$

$3b+6c \neq 2b+6c$

therefore

Choice A is not equivalent to the given expression

case B) (b+3c)+(b+3c)

Combine like terms

$b+3c)+(b+3c)=(b+b)=(3c+3c)=2b+6c$

$2b+6c= 2b+6c$

therefore

Choice B is equivalent to the given expression

case C) 2(b)+2(3c)

Multiply both terns by 2

$2(b)+2(3c)=2b+6c$

$2b+6c= 2b+6c$

therefore

Choice C is equivalent to the given expression

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3 years ago

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Show a photo

Step-by-step explanation:

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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

2 years ago