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nata0808 [166]
3 years ago
11

The area of a circle is 144π in². What is the circumference, in inches? Express your answer in terms of π

Mathematics
1 answer:
Fofino [41]3 years ago
5 0

Answer:

42.54 inches

Step-by-step explanation:

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The volume of clay used to build a cylindrical pillar with a height of 9 centimeters is 324π cubic centimeters. The area of the
Kitty [74]
The volume of a cylinder is written as V = πr²h. The volume is 324π cm³ and the height is 9 cm. Evaluating it, we will arrive at the value of radius as 6 cm. Area of the base can be obtained using the formula A = πr². If r = 6cm, area of the base will be 36π cm
3 0
3 years ago
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one year farmers recieved an average of $13.145 per bushel of wheat. How much did the farmer receive for selling 100 bushels of
djyliett [7]

multiply the 2 numbers together

100 x 13.145 = $1,314.50

3 0
2 years ago
A circle has a radius of 2 meters. find the exact circumference of this circle in terms of pi
VLD [36.1K]

Answer:

See below.

Step-by-step explanation:

The circumference of a circle is 2\pi r

We know that the radius is 2 meters, so:

2(2) \pi

4\pi meters.

5 0
3 years ago
1.A pizza place offers 6 different cheeses and 12 different toppings. In how many ways can a pizza be made with 1 cheese and 3 t
NNADVOKAT [17]
1.) You have 12 toppings. You choose one topping--that leaves you with 11 toppings. You choose another--that leaves 10. 12×11×10 = 1,320.
Multiply 1,320 topping choices by 6 cheeses to get 7,920 total combinations.

2.) (Though I'm less sure of these)
CDs: 6×5×4×3×2 = 720
Cassettes: 5×4 = 20
DVDs: 8×7×6×5 = 1680
4 0
3 years ago
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}}

(c) Parameterize C by

\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
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