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olya-2409 [2.1K]
3 years ago
10

-21 = 4x + 7 It's those 2 step equations please explain how to do it thank you.

Mathematics
1 answer:
nordsb [41]3 years ago
8 0

Answer:

-21-7=4x

-28=4x

-28/4=x

-7=x

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Rond to the nearest cents 0.5239 and 32,048,87219
geniusboy [140]

Answer:

0.5239 would round to 0.52

3,204,887,219 would round to 3,000,000,000

Step-by-step explanation:


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Write the following in decimal format:<br> 364<br> 1000
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2 years ago
A rectangular box has interior dimensions 6-inches by 5-inches by 10-inches. The box is filled with as many solid 3-inch cubes a
Arte-miy333 [17]

Answer: The box would have 99% of its volume taken up.

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Length = 6 inches

Width = 5 inches

Height = 10 inches

Therefore the volume of the box shall become

Volume = L x W x H

Volume = 6 x 5 x 10

Volume = 300 cubic inches

Also a 3 inch cube would have its volume given as follows (

Volume = 3 x 3 x 3 (All sides of a cube has equal lengths)

Volume = 27 cubic inches

To find out how many of 3-inch cubes can fit in, divide 300 by 27 and that equals 11.11.

Hence you can have at most 11 cubes in the box. The total volume of 11 cubes is given as 11 x 27 which equals 297. Therefore, the percentage of the box taken up completely by the cubes is given as;

Percentage = (Volume of cubes/Volume of box) x 100

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Therefore the box would have 99% of its volume taken up by the cubes.

8 0
3 years ago
If lim x-&gt; infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
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We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
2 years ago
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