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olga_2 [115]
2 years ago
9

Does anyone know this? I’m really stuck

Mathematics
2 answers:
S_A_V [24]2 years ago
7 0

<em>p = 21</em>

Step-by-step explanation:

When you have a midsegment, the base is going to be two times as large as the midsegment. You can find the midsegment if you multiply it by two and set it equal to the base. I have attached a photo below that may help you.

This is the equation we will be using.

2(p-13)=p-5

Now let's solve! Let's start by distributing the 2 by what's in the parentheses.

2p-26=p-5

Subtract p from both sides.

p-26=-5

Add 26 to both sides.

<u>p = 21</u>

Viktor [21]2 years ago
7 0
P = 21 by the angles. please mark me as brainliest
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1. I suppose the ODE is supposed to be

\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)

Solving for \dfrac{\mathrm dy}{\mathrm dt} gives

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\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}

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\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)

On the right, we have

\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)

\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C

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2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C

\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C

y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}

y^{1/2}=C(1-t^2)^{1/4}-1

I'll leave the solution in this form for now to make solving for C easier. Given that y\left(-\dfrac12\right)=1, we get

1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1

2=C\left(\dfrac54\right)^{1/4}

C=2\left(\dfrac45\right)^{1/4}

and so our solution is

\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}

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