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3 years ago

What is the area of the following circle r=2

Mathematics

1 answer:

denpristay [2]3 years ago

8 0

Answer:

12.57

Step-by-step explanation:

$\pi r^2\\\pi (2)^2\\=12.57$

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Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

1 year ago

Which expression is equivalent to the following complex fraction?

Virty [35]

Answer:

$\dfrac{-4x+7}{2(x-2)}$

Step-by-step explanation:

Write the terms of numerator and denominator using a common denominator.

$\dfrac{\dfrac{3}{x-1}-4}{2-\dfrac{2}{x-1}}=\dfrac{\left(\dfrac{3-4(x-1)}{x-1}\right)}{\left(\dfrac{2(x-1)-2}{x-1}\right)}}=\boxed{\dfrac{-4x+7}{2(x-2)}}$

7 0

3 years ago

Read 2 more answers

What is the prime factorization of 68

umka21 [38]

68 is not a prime number. The prime factorization of 68 would be 2 x 2 x 17.

6 0

3 years ago

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Which type of graph would display the cilnic totals for each of the 12 months?

BartSMP [9]

the dot plot, because box plot only shows the 5 number summary and histograms show ratios

3 0

3 years ago

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10. The perimeter of a parallelogram is 72 meters. The width of the

nalin [4]

Answer:

Length is 20 meters and Width is 16 meters

Step-by-step explanation:

A parallelogram is a 4-sided figure with 2 pair of parallel sides.

The perimeter is the sum of all 4 sides of the parallelogram.

Let the length of parallelogram be x and width be y.

The perimeter is 72, thus we can write:

x + x + y + y = 72

2x + 2y = 72

Also, width is 4 less than length, thus we can write:

y = x - 4

Now, we replace the 1st equation with the 2nd equation and solve for x first. Shown below:

2x + 2y = 72

2x + 2(x - 4) = 72

2x + 2x - 8 = 72

4x = 72 + 8

4x = 80

x = 80/4 = 20

And now y:

y = x - 4

y = 20 - 4

y = 16

Thus, the Length is 20 meters and Width is 16 meters

4 0

3 years ago