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liraira [26]
3 years ago
10

Write an equation for the parabola that passes through (-2, 7) , (1, 10) , and (2, 27) .

Mathematics
1 answer:
vlada-n [284]3 years ago
5 0
(-2,7) (1,10) (2,27)
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Alice found the area of the figure by breaking apart it apart where the dashed line is drawn. Her work is shown below. area of r
Alex Ar [27]

Answer:

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Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

We know the game and we're gonna play it

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Never gonna give you up, never gonna let you down

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Never gonna give you up, never gonna let you down

Never gonna run around and desert you

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Step-by-step explanation:

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3 years ago
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P = {52, 77, 91, 124,<br> 217)<br> Three members of the set P have a common<br> factor which is
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tis always a good idea when factoring to start off with a quick prime factoring.

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4 0
2 years ago
−4 = 2t − 2 what does t equal
sladkih [1.3K]

Answer:

t = -1

Step-by-step explanation:

-4 = 2t - 2

2t = -2

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3 years ago
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adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

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3 years ago
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Goryan [66]
5x+20+16
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