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arlik [135]
3 years ago
9

6 and 3/4 cups of fruit salad was made. They ate 1/3 of the salad. How much fruit salad did they eat?

Mathematics
1 answer:
andrezito [222]3 years ago
8 0

Answer:

6 1/2

Step-by-step explanation:

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In 2016, the top 1 percent of taxpayers accounted for more income taxes paid than the bottom 90 percent combined. The top 1 perc
qaws [65]

Answer:

$464Billion

Step-by-step explanation:

If 37.3%=$538B

32.2%=?

By \ Cross \ Multiplying:-\\32.2\%=>\frac{32.2\%*538}{40\%}\\=464.440

32.2% is roughly $464Billion in taxes

3 0
3 years ago
Help me to solve problem pless​
sergejj [24]

Answer:

a:

5 x -3 +6

= 5 x 6 + -3

= 30 + -3

= 27

8 0
3 years ago
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The vending machine has sodas in a ratio of 3 Pepsi to 4 coke. If there are 24 Pepsis in the machine, how many total sodas are t
Natasha2012 [34]

Step-by-step explanation:

let the ration be c

3x and 4x

Now

3x +4x=24

4 0
3 years ago
Stan's lawn mower had ⅛ of a gallon of gasoline in the tank. Stand started mowing and used all of the gasoline. He put 6/10 of a
KATRIN_1 [288]

Answer:Stan used 19/40 of a gallon of gasoline.

Step-by-step explanation:

At first, Stan's lawn mower had 1/8 of a gallon of gasoline in the tank. When it got finished, he put 6/10 of a gallon of gasoline in the tank. It means that the number of gallons of gasoline that he has already put in the tank is

1/8 + 6/10 = 29/40 of a gallon.

After he mowed, 1/4 of a gallon was left in the tank. Therefore, the total amount of gasoline Stan used is

29/40 - 1/4 = 19/40 of a gallon

8 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
2 years ago
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