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2 years ago

13

Selena want s to find the area of the figure below. what is the area of the figure?

1 answer:

xenn [34]2 years ago

5 0

Answer:

52 m sq

Step-by-step explanation:

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Convert 5. 06km to metres

djyliett [7]

Answer:

5060

Step-by-step explanation:

3 0

2 years ago

If segment UV is considered to be the base of parallelogram TUVW, which segment would represent the altitude of the parallelogra

omeli [17]

The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>Segment TX is the segment that would best represent the altitude of the parallelogram.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

3 years ago

The area ( l x w ) of the postcard is 24 square inches. What is the width b of the message (in inches)?

pentagon [3]

area = (l)(w)(h)

24 = 4(3 + b)

24 = 12 + 4b

-12 on both sides

12 = 4b

/4 on both sides

b = 3 inches

Substitute b = 3 inches in width

3 + b

3 + 3 = 6 inches

Answer:

width = 6 inches

5 0

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:

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

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

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

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

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

4 0

2 years ago

A point is reflected across the x-axis. The new point is (5 -3.5) What is the distance between the two points?

alukav5142 [94]

Answer:

10

Step-by-step explanation:

8 0

2 years ago