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Dmitrij [34]
2 years ago
10

Opposite in meaning

Mathematics
1 answer:
ICE Princess25 [194]2 years ago
6 0

1 (hide)

2 (disattracted)

3(...)

4(younger)

5(decided)

6(big)

7(wide)

8(down)

9(stopped)

10(

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How do i draw two number lines that show 0.20 and 1/5 are equivalent?
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The frist line may be one with the numbers 0, 1 and 2 and 10 divisions (marks at equal distance) between each integer. Each division will be equal to 0.1 units and then you can mark the second division from the zero point to the right as the 0.20 mark.

The other line must have the same integers, 0 , 1 and 2 placed in identical form as the first line. Then
   - draw an inclined straight line since the point zero,
   - mark 5 points in the inclined lined equally spaced over the line.
   - draw a sttraight line from the 5th point to the point with the mar 1 over the base number line.
   -  draw a parallel line to the previous one passing trhough the second point of the inclined line and mark the point where this parallel touchs the base number line. This point shall be at the same distance from zero than the 0.2 mark was in the first number line, meaning that 0.2 and 1/5 are equivalent.
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2 years ago
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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Answer:

I think it is A

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