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3 years ago

What is the gcf between 96 and 48

Mathematics

2 answers:

Andru [333]3 years ago

5 0

The gcf is "greatest common factor". You can factor 48 from both 48 and 96, (2x48=96), thus your gcf is 48.

coldgirl [10]3 years ago

4 0

<span>Your answer is: 2 x 2 x 2 x 2 x 3 = 48.</span>

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The number of items (N) that can be purchased for a given amount of money is inversely proportional to the cost (C) of an item.

mars1129 [50]

Answer:

436 items

Step-by-step explanation:

When a quantity A is inversely proportional to another quantity B, we can write:

$A=\frac{k}{B}$

Where k is the proportionality constant [that we need to find]

For this problem, we can write the ratio as:

$N=\frac{k}{C}$

Now given N = 480 and C = 4.90, we first find k:

$N=\frac{k}{C}\\480=\frac{k}{4.90}\\k=4.90*480\\k=2352$

Now, as second step, we need this time to find N, when

C = 5.40

Note, we know know k = 2352, so we find N:

$N=\frac{k}{C}\\N=\frac{2352}{5.40}\\N=435.55$

Rounding to next whole number, we have 436 items

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3 years ago

How to turn a fraction into a division problem?

V125BC [204]

X/y = x divided by y

3 0

3 years ago

Someone help me out please

MA_775_DIABLO [31]

Answer:

142.5

Step-by-step explanation:

π×11²×135/360

= 363π/8

= 142.5 (rounded to the nearest tenth)

Answered by GAUTHMATH

6 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

$\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}$

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2 years ago

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WINSTONCH [101]

Answer:

Step-by-step explanation:

check the attached files below

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4 years ago