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leva [86]
3 years ago
12

PLEASE HELP ME IM SO CONFUSED The volume of a cube is x³ cubic units. The surface area of the cube is x³ square units. What is t

he value of x?
Mathematics
1 answer:
Tom [10]3 years ago
3 0

Answer: The value of x is 6.

Step-by-step explanation:

TO find the volume of a cube you cube any side length. so if the side  length is 6.

6^3 = 216

To find the surface area of a cube you will square one of the side lengths and multiply it by 6.

so the side length is 6 and 6 squared is 36

36*6 = 216

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Write an equivalent expression for 4g+2 using the distributive property.
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4 * g + 2 ....... or do u want me to figure what g is 
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Indira finds that on a typical day, 4 out of every 5 students at her school eat a
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396

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it's it's 396 because if you add 99 more you will get 495.

each percentage is 99, so 99 x 5 is how many students there are 99 x 4 is the answer.

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A mechanic gave a parts store a deposit of $25 toward the purchase of a part costing $89.95. How much does the mechanic still ow
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3 0
3 years ago
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Simplify: x(x-2)-3x(2x-1)
julsineya [31]

Answer:

-5x^2 +x

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x(x-2)-3x(2x-1)

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x^2 -2x -6x^2 +3x

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4 0
3 years ago
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This is a question on my partial fractions homework, but no matter what I try I can't figure it out..
Ierofanga [76]
\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}
\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}
\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)
\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3

So you have

\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}
=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}

where in the first integral we substitute x\mapsto x+1.

=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}
=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)
=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2
=\dfrac23+\ln\dfrac89
4 0
3 years ago
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