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2 years ago

Joy is 5 years older than Kim. This can be modeled with the following equation.

Mathematics

2 answers:

klasskru [66]2 years ago

3 0

K=J-5 Because kim is 5 years younger than joy.

romanna [79]2 years ago

3 0

K=j-5 so thats what the form would be

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3 years ago

Whats the equation for <br>4n + 6 - 2n= 2 (n+3)

lisabon 2012 [21]

4n + 6 - 2n = 2(n + 3)

Simplify.

4n + 6 - 2n = 2n + 6

Subtract 6 from both sides.

4n - 2n = 2n + 6 - 6

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4n - 2n - 2n = 6 - 6

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Therefore, there are infinite solutions.

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3 years ago

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3 years ago

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The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas

iren [92.7K]

For the last part, you have to find where $f'(t)$ attains its maximum over $0\le t\le8$ . We have

$f'(t)=-6\sin3t+6\cos2t$

so that

$f''(t)=-18\cos3t-12\sin2t$

with critical points at $t$ such that

$-18\cos3t-12\sin2t=0$

$3\cos3t+2\sin2t=0$

$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

$\cos t(12\sin^2t-4\sin t-3)=0$

So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$ , and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$ . They are approximately