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Ray Of Light [21]
3 years ago
7

Im working on reflections and I am confused on what the reflection of A would be? I'm trying to find the reflection of (0,5) ove

r the Yaxis? The rule is this: (x,y)->(-x,y) but there is no such thing as -0......so... yea thats why im confused? Or is the reflection unchanging? is A'=0,5?

Mathematics
1 answer:
Oduvanchick [21]3 years ago
7 0
Any points that lie on the line of reflection (the mirroring line) will not move. So point A = (0,5) will become A' = (0,5) staying in the same location. Anything on the y axis will stay where it is.
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Solve the initial value problem
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9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}

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\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}

Multiply both sides by \mu:

(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}

Condense the left side as the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}

Integrate both sides:

(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt

For the integral on the right, substitute

u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du

\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du

\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C

\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)

\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C

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Given that y(0)=12, we get

12=9+C\implies C=3

\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}

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