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dsp73
3 years ago
15

Which right prism would have the same volume as a square prism with a base area of 36 m2 and a height of 3 m?

Mathematics
2 answers:
Neko [114]3 years ago
8 0
Well to find the volume of the square prism, first you need to multiply the base area (36) by its height (3) and you would end up with 108m3 as the volume. You may not be familiar with this because u are most likely taught length times width times height as volume, but what i just did is the same because length times width is the sum of the base area, and height was already given. Anyways, now that we have gotten the square prism volume (108m3) we need to find which prism has the same volume. I’m gonna save u from searching and tell you that it is the rectangular prism because length times width times height (18x2x3) equaled 108 just like the square prism.
Anna007 [38]3 years ago
7 0

Answer:

the answer is 18x2x3

Step-by-step explanation:

i took the test

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The answer would be 1,2,4,3

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We know that he is 6 songs away from 100.
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3 years ago
I need help with this
larisa [96]

Answer:

Not similar

Step-by-step explanation:

Even though

8/2=16/4,

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3 years ago
Because of their connection with secant​ lines, tangents, and instantaneous​ rates, limits of the form ModifyingBelow lim With h
Gre4nikov [31]

Answer:

\dfrac{1}{2\sqrt{x}}

Step-by-step explanation:

f(x) = \sqrt{x} = x^{\frac{1}{2}}

f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}

We use binomial expansion for (x+h)^{\frac{1}{2}}

This can be rewritten as

[x(1+\dfrac{h}{x})]^{\frac{1}{2}}

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}

From the expansion

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots

Setting x=\dfrac{h}{x} and n=\frac{1}{2},

(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots

=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots

Multiplying by x^{\frac{1}{2}},

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots

The limit of this as h\to 0 is

\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}} (since all the other terms involve h and vanish to 0.)

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The answer is y=4-x.
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Read 2 more answers
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