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3 years ago

Which line segments do you draw to find the center of a circle circumscribed about a triangle?

1 answer:

6 0

You draw perpendicular bisectors to find the center of a circle circumscribed about a triangle. Hook me up with 5 stars :)

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Simplify<br> (-4-8i)-(-1-3i)

AfilCa [17]

3 - 2i is the answer .

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3 years ago

Solve the equation.

Ad libitum [116K]

Answer:

D. 13 1/2

Step-by-step explanation:

2/3 x -2 = 7

2/3 x -2 + 2 = 7 + 2

2/3 x = 9

x = 9 (3/2)

x = 27/2

x = 13 1/2

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3 years ago

Help please will award brainliest!!

OleMash [197]

Answer:

The answer is -x+12/ x (x-2)

Step-by-step explanation:

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2 years ago

Read this statement: “If a triangle is isosceles, then it is equilateral.” What is the converse of the statement?

makkiz [27]

From the above statement that <span>“If a triangle is isosceles, then it is equilateral" the only thing i can conclude is </span>If a triangle is equilateral, then it is isosceles.

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3 years ago

Read 2 more answers

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:

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

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

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

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

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

4 0

2 years ago