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3 years ago

How to find the area of 2 half circles and a rectangle

1 answer:

6 0

Hello!

We use different formulas to calculate the areas of different shape.

RECTANGLE:

To find the area of a rectangle, we must simply multiply its length by its width. The formula for its area is:

A = l × w

SEMICIRCLE:

Since the formula for a circle is pi × r × r, we must use the same formula but divide it in half, because a semicircle is a half circle, which is why its area would also be half of a circle's. The formula for a semicircle's area is:

A = 1/2 pi × r × r

Tip:

Write these formulas down and memorize them so that you don't forget them. You'll have to use these formulas quite often when finding the area of these shapes.

You might be interested in

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

2 years ago

Roots of x^2+8x+9 by finding in quadratic formula

gladu [14]

The quadratic formula is -b + or - the square root of b squared - 4 times the a value and c value over 2a. So the roots would be -1.354249 and -6.645751. I believe these are the roots.

4 0

3 years ago

How does tan 135 degrees compare to tan -135 degrees

Anna007 [38]

Evaluating the tangent function for the two values we obtain the following:
tan 135=tan(135+180)=tan 315=-1

next:
tan -135=tan(180-135)=tan 45=1
thus comparing the two we see that tan 135 is negative while tan -135 is positive

3 0

3 years ago

$4750 invested at 10% p.a compounded semiannually for 10 years, calculate the interest

alexgriva [62]

<h3>Answer: $7,853.16</h3>

Work Shown:

A = P*(1+r/n)^(n*t)

A = 4750*(1+0.10/2)^(2*10)

A = 12,603.164099436

A = 12,603.16

That represents the account balance after 10 years. To find the interest only, subtract off the deposit amount.

interest = A-P = 12,603.16 - 4,750 = 7,853.16

6 0

2 years ago

hi im a kid named tyler uzing mi big broters acont to ask this qwestion. wats 46+13? tanks for the help

amm1812

Answer:

Step-by-step explanation:

ur welcome

6 0

2 years ago