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3 years ago

9

Please help, I will really appreciate all your help please :(

1 answer:

Grace [21]3 years ago

8 0

Answer:

should be y=-2/1x+2

Step-by-step explanation:

because the slope is going down 2 over 1 and the y-intercept is at positive 2

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Perform the indicated operation (-25) - (25)

slavikrds [6]

Hello,

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3 years ago

Five higschool friends all happen to be female born in 2003 assuming that the survival of each ine is independent what is the pr

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Step-by-step explanation:

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Anna11 [10]

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2 years ago

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Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

zloy xaker [14]

Answer:

$y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

$x={\pi}{4}$

Step-by-step explanation:

We are given that

$y'=\frac{2cos2x}{3+2y}$

y(0)=-1

$\frac{dy}{dx}=\frac{2cos2x}{3+2y}$

$(3+2y)dy=2cos2x dx$

Taking integration on both sides then we get

$\int (3+2y)dy=2\int cos 2xdx$

$3y+y^2=sin2x+C$

Using formula

$\int x^n=\frac{x^{n+1}}{n+1}+C$

$\int cosx dx=sinx$

Substitute x=0 and y=-1

$-3+1=sin0+C$

$-2=C$

$sin0=0$

Substitute the value of C

$y^2+3y=sin2x-2$

$y^2+3y-sin 2x+2=0$

$y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

Hence, the solution $y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

When the solution is maximum then y'=0

$\frac{2cos2x}{3+2y}=0$

$2cos2x=0$

$cos2x=0$

$cos2x=cos\frac{\pi}{2}$

$cos\frac{\pi}{2}=0$

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

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3 years ago

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adoni [48]

Answer: it’s C

Step-by-step explanation:

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3 years ago

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