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3 years ago

A can holds 2 liters of paint.then how many can we fill with 11 liters of paint

Mathematics

1 answer:

pychu [463]3 years ago

4 0

Answer:

5 and a half

Step-by-step explanation:

Because you can fit 2 liters in each can, to find how many you would need, divide 11 by 2, this would give you 5 and a half, 5.5.

It could be that they want you to round up to six or round down to five, so do consider that.

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Luba_88 [7]

Answer:

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Step-by-step explanation:

The mode of 44 and 51 is 47.50 and 47.50 x 6 equals 285

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3 years ago

Please help!!!!!!!!!!!!!!!!!!!!!

3241004551 [841]

Answer:

C and D

Step-by-step explanation:

The arc length is calculated as

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The area (A) of a sector is calculated as

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3 years ago

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3 years ago

Read 2 more answers

Given the function f ( x ) = 5 + x 2 , calculate the following values:

Drupady [299]

Answer:

f(x+1)=fx+f

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Step-by-step explanation:

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2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E2%20%5Cleft%28%20%5Csqrt%5B3%5D%7B%20%7Bx%7D%5E%7B2%7D%20%20%2B%20

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If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show

$\displaystyle \int_a^b f(x) \, dx = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) \, dx$

Let $f(x) = \sqrt{1 + x^3}$ . Compute the inverse:

$f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}$

and we immediately notice that $f^{-1}(x+1)=\sqrt[3]{x^2+2x}$ .

So, we can write the given integral as

$\displaystyle \int_0^2 f^{-1}(x+1) + f(x) \, dx$

Splitting up terms and replacing $x \to x-1$ in the first integral, we get

$\displaystyle \int_1^3 f^{-1}(x) \, dx + \int_0^2 f(x) \, dx = 2 f(2) - 0 f(0) = 2\times3+0\times1=\boxed{6}$

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2 years ago

A can holds 2 liters of paint.then how many can we fill with 11 liters of paint​

A can holds 2 liters of paint.then how many can we fill with 11 liters of paint