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2 years ago

Please help. Questions 5-9

Mathematics

1 answer:

Salsk061 [2.6K]2 years ago

3 0

5. Not true/can’t be solved
6. t=62
7. X=-6
8. X=5
9. $4.72 per student

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From a practice assignment:<br>solve the following differential equation given initial conditions

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If $y' = e^y \sin(x)$ and $y(-\pi)=0$ , separate variables in the differential equation to get

$e^{-y} \, dy = \sin(x) \, dx$

Integrate both sides:

$\displaystyle \int e^{-y} \, dy = \int \sin(x) \, dx \implies -e^{-y} = -\cos(x) + C$

Use the initial condition to solve for $C$ :

$-e^{-0} = -\cos(-\pi) + C \implies -1 = 1 + C \implies C = -2$

Then the particular solution to the initial value problem is

$-e^{-y} = -\cos(x) - 2 \implies e^{-y} = \cos(x) + 2$

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1 year ago

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2 years ago

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BigorU [14]

$\bf{-8x+4\leq 36 }$

Subtract 4 from both sides.

$\bf{-8x\leq 36-4 }$

Subtract 4 from 36 to get 32.

$\bf{-8x\leq 32 }$

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$\bf{x\geq \dfrac{32}{-8} }$

Divide 32 by −8 to get −4.

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<h2>{ Pisces04 }</h2>

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1 year ago

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Answer:

i think the 2nd one

Step-by-step explanation:

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