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3 years ago

13

Please help me :( on this

1 answer:

sashaice [31]3 years ago

7 0

Answer:

A. point F (3,5) B. 5 units C. 10 units

Step-by-step explanation:

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Retire zero la bour li dan trou fess to pu ggnr bon

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4 years ago

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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3 years ago

9. Cassie bought 15 tubes of oil paint for her art class. The paint cost $37.35 altogether.

elena55 [62]

Answer:Multiplication

Step-by-step explanation:

It’s just 15.00 times 37.35

3.5 is to throw u off

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Step-by-step explanation:

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