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3 years ago

What is the volume of a circle

Mathematics

2 answers:

Vikentia [17]3 years ago

4 0

Answer:

Calculate the area of the base (which is a circle) by using the equation πr² where r is the radius of the circle. Then, multiply the area of the base by the height of the cylinder to find the volume.

Circumference = 2 • π • radius = π • diameter

Circle Area = π • r² = ¼ • π • d²

The circumference of a circle is directly proportional to its radius. The ratio between a circle's circumference and its diameter is pi, a constant that equals approximately 3.142. The circle's diameter, in turn equals twice the radius. A regular solid that has a circle as a face is called a cylinder. Like all geometric prisms, a cylinder's volume is the product of its cross-sectional area and its length. The cross-sectional area is the area of its circle, and you can calculate it from the circle's radius. Multiply the circle's radius by 2 to calculate its diameter. For instance, if the radius equals 6 inches, compute as follows to determine the diameter: 6 x 2 = 12 inches. Multiply the diameter by pi to get the circle's circumference: 12 x 3.142 = approximately 37.7 inches. Find the square of the radius: 6 x 6 = 36 square inches. Multiply the answer determined in Step 3 by pi to get the circle's area: 36 x pi = 113 square inches. Multiply the circle's area by the cylinder's length to obtain the volume. If the length is, for instance, 10 inches, then compute as follows: 113 x 10 = 1,130 cubic inches.

Ymorist [56]3 years ago

3 0

Answer:

volume = 4/3 × pi × r^3

Step-by-step explanation:

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Sin(x)^4+ cos(x)^4=1/2

posledela

I'll assume that what was meant was $\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$ .

The exponent in the funny place is just an abbreviation: $\sin ^4 x = (\sin x)^4$ .

I hope that's what you meant. Let me know if I'm wrong.

Let's start from the old saw

$\cos^2 x + \sin ^2x = 1$

Squaring both sides,

$(\cos^2 x + \sin ^2x)^2 = 1^2$

$\cos^4 x + 2 \cos ^2 x \sin ^2x +\sin ^4x = 1$

$\cos^4 x + \sin ^4x = 1 - 2 \cos ^2 x \sin ^2x$

So now the original question

$\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$

becomes
$1 - 2 \cos ^2 x \sin ^2x = \dfrac{1}{2}$

$4 \cos ^2 x \sin ^2x = 1$

Now we use the sine double angle formula

$\sin 2x = 2 \sin x \cos x$

We square it to see

$\sin^2 2x = 4\sin^2 x \cos^2 x = 1$

Taking the square root,

$\sin 2x = \pm 1$

Not sure how you want it; we'll do it in degrees.

When we know the sine of an angle, there's usually two angles on the unit circle that have that sine. They're supplementary angles which add to $180^\circ$ . But when the sine is 1 or -1 like it is here, we're looking at $90^\circ$ and $-90^\circ$ , which are essentially their own supplements, slightly less messy.

That means we have two equations:

$\sin 2x = 1 = \sin 90^\circ$

$2x = 90^\circ + 360^\circ k \quad$ integer $k$

$x = 45^\circ + 180^\circ k$

or

$\sin 2x = -1 = \sin -90^\circ$

$2x = -90^\circ+ 360^\circ k$

$x = - 45^\circ + 180^\circ k$

We can combine those for a final answer,

$x = \pm 45^\circ + 180^\circ k \quad$ integer $k$

Check. Let's just check one, how about

$x=-45^\circ + 180^\circ = 135^\circ$

$\sin(135)= 1/\sqrt{2}$

$\sin ^4(135)=(1/\sqrt{2})^4 = 1/4$

$\cos ^4(135)=(-1/\sqrt{2})^4 = 1/4$

$\sin ^4(135^\circ) +\cos ^4(135^\circ) = 1/2 \quad\checkmark$

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Answer:

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Step-by-step explanation:

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