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lianna [129]
3 years ago
12

Find all points having an x-coordinate of 2 whose distance from the point (-1,-2) is 5

Mathematics
1 answer:
mariarad [96]3 years ago
5 0
The\ equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\where\ (a;\ b)\ -the\ coordinates\ of\ the\ center;\ r-the\ radius\\-----------------------------\\(-1;-2)-the\ center\ of\ the\ circle\\r=5\\\\The\ equation:(x+1)^2+(y+2)^2=5^5\\\\Put\ x=2\ to\ the\ equation:\\\\(2+1)^2+(y+2)^2=25\\3^2+(y+2)^2=25\\9+(y+2)^2=25\\(y+2)^2=25-9\\(y+2)^2=16\iff y+1=\pm\sqrt{16}\\\\y+1=-4\ or\ y+1=4\ \ \ \ \ |subtract\ 1\ from\ both \sides\\y=-5\ or\ y=3\\\\Answer:(2;-5)\ and\ (2;\ 3).
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\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}

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Plugging the values into the equation, we have

\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}

Taking Ln of both sides, we have

\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}

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1 year ago
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Step-by-step explanation:

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