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3 years ago

What is the solution to 5/6x-1/3>1 1/3

Mathematics

1 answer:

CaHeK987 [17]3 years ago

3 0

Answer:

Step-by-step explanation:

$\frac{5}{6}x-1>1\frac{1}{3}\\\\\frac{5}{6}x-1>\frac{4}{3}\\\\\frac{5}{6}x>\frac{4}{3}+1\\\\\frac{5}{6}x>\frac{4}{3}+\frac{3}{3}\\\\\frac{5}{6}x>\frac{7}{3}\\\\x>\frac{7}{3}*\frac{6}{5}\\\\x>\frac{7*2}{5}\\x>\frac{14}{5}\\\\x>2\frac{4}{5}$

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Calculus 3 help please.

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

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

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

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

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(b) Parameterize $C$ by

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Integrate by parts with

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

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

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

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with $0\le t\le1$ . Then

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and

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