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2 years ago

Given the conversion factor 12 in/ 1 ft, which has the larger volume?

Mathematics

2 answers:

Marrrta [24]2 years ago

6 0

A is the answer. We know this because a factor of 12in / 1 ft has a lager volume

dsp732 years ago

5 0

Answer:

A. Cube A

Given:

Cube A volume = 5832 in³

For Cube B dimension in inches:

1.2 feet = 14.4 in

1 feet = 12 in

1 feet = 12 in

So Volume of Cube B:

L × B × H = 14.4 × 12 × 12 = 2073.6 in³

Cube A has a larger volume that Cube B.

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sineoko [7]

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3 years ago

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3 years ago

What do you get when u count out 40 dollars with quarters? help me please

Rudiy27

Answer:

160 Quarters

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3 years ago

Use the definition of the derivative as a limit to find the derivative f′ where f(x)= √ x+2.

MA_775_DIABLO [31]

Step-by-step explanation:

If the equation is

$\sqrt{x + 2}$

Then, here is the answer.

The definition of a derivative is

$\frac{f(x + h) - f(x)}{h}$

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So we get

$\frac{ \sqrt{x + h + 2} - \sqrt{x + 2} }{h}$

Multiply both equations by the conjugate.

$\frac{ \sqrt{x + h + 2} - \sqrt{x + 2} }{h} \times \frac{ \sqrt{x + h + 2} + \sqrt{x + 2} }{ \sqrt{x + h + 2} + \sqrt{x + 2} } = \frac{x + h + 2 - (x + 2)}{h \sqrt{x + h + 2} + \sqrt{x + 2} }$

$\frac{h}{h \sqrt{x + h + 2} + \sqrt{x + 2} }$

$\frac{1}{ \sqrt{x + h + 2} + \sqrt{x + 2} }$

Since h is very small, get rid of h.

$\frac{1}{ \sqrt{x + 2} + \sqrt{x + 2} }$

$\frac{1}{2 \sqrt{x + 2} }$

So the derivative of

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Part 2: If your function is

$\sqrt{x} + 2$

Then we get

$\frac{ \sqrt{x + h} + 2 - ( \sqrt{x} + 2) }{h}$

$\frac{ \sqrt{x + h} - \sqrt{x} }{h}$

$\frac{x + h - x}{h( \sqrt{x + h} + \sqrt{x}) }$

$\frac{h}{h( \sqrt{x + h} + \sqrt{x} ) }$

$\frac{1}{ \sqrt{x + h} + \sqrt{x} }$

$\frac{1}{2 \sqrt{x} }$

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3 0

3 years ago

Simpify by combing like terms -4x5x-17-22x simply

Gre4nikov [31]

Answer:

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Step-by-step explanation:

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6 0

3 years ago