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Sever21 [200]
3 years ago
7

The circumference of circle A is four times the circumference of circle B. Which statement about the areas of the circles is tru

e?
Mathematics
1 answer:
Brums [2.3K]3 years ago
3 0

Answer:

The area of circle A is 16 times the area of circle B.

Step-by-step explanation:

he area of circle A is three times the area of circle B.

It depends on the value of π.

It depends on the actual diameters of the circles.

The area of circle A is 16 times the area of circle

the circumference of a circle = 2 x pi x r

Let imagine that the radius of B is 2cm

then the circumference of B = 2 x 2 x pi = 4pi

If the circumference of a is 4 times that of B, it means the circumference is 16 and the radius is 8

Area of a circle is pi x r^2

Area of B = pi x 2^2 = 4pi

Area of A = pi x 8^2 = 64pi

64/4 = 16 pi

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siny=2.7sin63/2.8

y=arcsin((2.7sin63)/2.8)

y≈59.23°  (to nearest one-hundredth of a degree)
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Find the supplement of an angle that measures 20 degrees.
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Which of the following is another name for a proof by contradiction?
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Step-by-step explanation:

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3 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} }

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\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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