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3 years ago

What is the volume of the rectangular prism, when the height is 10 1/2, the width is 2 3/4 and the base is 5?

1 answer:

5 0

The formula for distinguishing the volume of a rectangular prism is the Base being multiplied by the Height. In other words, the formula is:
V = Bh

Now, we need to substitute our values for each variable to solve for our volume.
V = 5 x 10.5; multiply the two numbers.
V = 52.5.

Your answer for this is "C.) 52 1/2" (as "52.5" is equivalent to "52 1/2").

I hope this helps!

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Please help if you know the answer

worty [1.4K]

It's C. When you're dividing g(x) by f(x), you get the equations in both B. and C., but C is more correct because you can't have x=-1/3 because the function is undefined at that number (you can't have the denominator equal 0)

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2 years ago

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The 8 red markers in a package account for 4% of all markers in the package. How many markers are in the package?

Vlad [161]

Answer:

192 markers

Step-by-step explanation:

if 8 = 4% then 2=1%

so we can assume there are 200markers in the package.

8-200= 192 markers

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2 years ago

If the perimeter is 168 and the width is 6 feet longer than thr length what are the dimensions

Ksenya-84 [330]

$\bf \textit{perimeter of a rectangle}\\\\ P=2(L+w)~~ \begin{cases} L=length\\ w=width\\ ----------\\ P=168\\ w=\stackrel{\textit{6 feet longer than L}}{L+6} \end{cases}\implies 168=2(L+\stackrel{w}{L+6}) \\\\\\ 84=2L+6\implies 78=2L\implies \cfrac{78}{2}=L\implies \boxed{39=L} \\\\\\ w=39+6\implies \boxed{w=45}$

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2 years ago

Pls help asap thank you

Sati [7]

Use the Pythagorean theorem to find the height.

Height = sqrt(26^2-10^2)

Height = 24 cm

Volume = 20^2 x 24/3

Volume = 7,200 cm^3

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2 years ago

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If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

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2 years ago