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2 years ago

7

Why do I measure capacity

1 answer:

earnstyle [38]2 years ago

7 0

To find the maximum for something pertaining to a liquid or etc.

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(10x2 + 99x – 5) ÷ (x + 10) =

Gennadij [26K]

The answer is 15+99x
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x+10

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2 years ago

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The table below shows two equations:

ra1l [238]

The correct answer is option A A) Equation 1 and equation 2 have no solutions For the equation 1 l 3x-1 l +7=2 l 3x-1 l = 2-7 =-5 The absolute value should be positive So, there is no solution For the equation 2 l 2x+1 l +4 = 3 l 2x+1 l = 3 - 4= -1 The absolute value should be positive So, there is no solution

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Find the volume of the solid formed by revolving the region bounded by LaTeX: y = \sqrt{x} y = x and the lines LaTeX: y = 1 y =

Strike441 [17]

Answer:

The volume is:

$\displaystyle\frac{37\pi}{10}$

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

$\displaystyle\int_a^b\pi r^2dx$

Notice here the radius of the washer is the difference of the given curves:

$x-\sqrt{x}$

So the integral becomes:

$\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx$

We solve it:

Factor $\pi$ out and distribute the exponent (you can use FOIL):

$\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx$

Notice: $x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}$

So the integral becomes:

$\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx$

Then using the basic rule to evaluate the integral:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4$

Simplifying a bit:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4$

Then plugging the limits of the integral:

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Taking the root (rational exponents):

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Then doing those arithmetic computations we get:

$\displaystyle\frac{37\pi}{10}$

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3 years ago

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Answer:

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lidiya [134]

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