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3 years ago

Circle A’s circumference is 3 times the value of Circle B’s Area. Find Circle B’a radius.

Mathematics

1 answer:

Natali [406]3 years ago

5 0

(side note: your formula C = 2πr^2 is incorrect, it should be C = 2πr)

Answer:

r = 9

Step-by-step explanation:

First let's find the circumference of Circle A. To use the formula we need radius. Since we have a diameter just divide by 2 to get radius. 27 / 2 = 13.5

Now plug that into C = 2πr: 2π*(13.5) = 27π

The question says that the area of circle B is 3x this value, so Circle B's area must be: 27π * 3 = 81π

Now plug that into A = πr^2 and solve for r:

$81\pi=\pi r^{2}\\81=r^{2}\\\sqrt{81}=r\\9=r$

(edited to correct a brain fart)

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Dennis_Churaev [7]

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Step-by-step explanation:

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<em>Comment on area and volume in general</em>

You will note the presence of the factor πr² in these formulas. This is the area of the circular base of the object. That is, the volume is the product of the area of the base and the height. In general terms, ...

V = Bh . . . . . for an object with congruent parallel "bases"

V = (1/3)Bh . . . . . for a pointed object with base area B.

This is the case for any cylinder or prism, even if the parallel bases are not aligned with each other. (That is, it works for oblique prisms, too.)

Note that the cone, a pointed version of a cylinder, has 1/3 the volume. This is true also of any pointed objects in which the horizontal dimensions are proportional to the vertical dimensions*. (That is, this formula (1/3Bh), works for any right- or oblique pyramid-like object.)

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3 years ago

∫(cosx) / (sin²x) dx

kirza4 [7]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2822772

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx}\\\\\\
=\mathsf{\displaystyle\int\! \frac{1}{(sin\,x)^2}\cdot cos\,x\,dx\qquad\quad(i)}$

Make the following substitution:

$\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}$

and then, the integral (i) becomes

$=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\
=\mathsf{\displaystyle\int\! u^{-2}\,du}$

Integrate it by applying the power rule:

$\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\
\mathsf{=\dfrac{u^{-1}}{-1}+C}\\\\\\
\mathsf{=-\,\dfrac{1}{u}+C}$

Now, substitute back for u = sin x, so the result is given in terms of x:

$\mathsf{=-\,\dfrac{1}{sin\,x}+C}\\\\\\
\mathsf{=-\,csc\,x+C}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus</em>

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3 years ago