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11111nata11111 [884]
3 years ago
9

How many 2/3s are in 2?

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
7 0

Answer:

3

Step-by-step explanation:

2/3+2/3+2/3=2

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Stells [14]

The first one is -8+2 = -6

And the second one is 2 x 2 = 4

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AT&T offers a deal for internet service that is $20 a month. There is an intial fee of $50. If Tim spent $330 on his interne
GenaCL600 [577]

Answer:

162 months

Step-by-step explanation:

if I'm correct it should be 162

3 0
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Question
puteri [66]

Answer:

The salt concentration of the mixed solution = 10%

Step-by-step explanation:

First assuming that the concentration of salt given is on a vol% basis.

First solution has 0.7 gallon of 5% salt

Amount of salt in this solution = 5% × 0.7 = 0.035 gallon

Second solution has 0.3 gallon of 22% salt

Amount of salt in this solution = 22% × 0.3 = 0.066 gallon

Total amount of salt in the resulting solution = 0.035 + 0.066 = 0.101 gallon

Total volume of the resulting solution = 0.7 + 0.3 = 1.00 gallon

Concentration of the mixture of salt = (0.101/1) = 0.101 = 10.1% = 10% to the nearest percent

Hope this Helps!!!

3 0
3 years ago
Richard receives a 3.5% commission for selling cleaning supplies. What is his commission on sales of $122,591?
konstantin123 [22]

Answer:

We have to find 3.5% $122,591

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122591*3.5/100=4,290.685

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5 0
3 years ago
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Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
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