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Olegator [25]
4 years ago
8

4. How many terms are ano a. 3x+y-7

Mathematics
1 answer:
eimsori [14]4 years ago
7 0

Answer:

you cant add them together..

and what does are ano even mean-

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Which of the following represents a fraction that converts to a repeating decimal with a value less than one?
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Solve the initial value problem
tigry1 [53]

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)}

Look for an integrating factor \mu(t):

\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

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

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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