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3 years ago

If 1 liter = 1.057 quarts, how many quarts are in 1 kiloliter

2 answers:

7 0

If 1 liter = 1.057 quarts, how many quarts are in 1 kiloliter?
1KL X (1L/1000KL) X (1.057quarts/1L) <<< use a calculator
I hope I helped! :)

Mrrafil [7]3 years ago

5 0

Answer:

There are 1057 quarts in 1 kiloliter.

Step-by-step explanation:

Consider the provided statement.

It is given that 1 liter = 1.057 quarts

Multiply 1000 to the both side of the equation 1 liter = 1.057 quarts

1×1000 liter = 1.057×1000 quarts

1000 liter = 1057 quarts

1 kiloliter = 1057 quarts (∴ 1 kiloliter = 1000 liter)

Hence, there are 1057 quarts in 1 kiloliter.

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3 years ago

What expression is equivalent to 6+3(n-4)-8+2n

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Step-by-step explanation:

Comment

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2 years ago

How much is the number 727731000 in percent form

tankabanditka [31]

Answer:

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Step-by-step explanation:

"Percent" means "per 100" or "over 100". So, to convert 727731000 to percent we rewrite 727731000 in terms of "per 100" or over 100.

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Simplified Conversion:

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3 years ago

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Math question

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

$h = \frac{V}{\pi\cdot r^{2}}$

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

$h = \frac{V}{\pi\cdot r^{2}}$

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

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Step-by-step explanation:

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