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3 years ago

There are between 24 and 40 students in a class.

Mathematics

1 answer:

nexus9112 [7]3 years ago

7 0

There are 12 boys and 21 girls for a total of 33 students

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X – 4 = 2x + 9 help me solve it step by step !

VashaNatasha [74]

Step-by-step explanation:

x - 4 = 2x + 9

x - (2x) = 9 + (4)

-x = 13

x = -13.

8 0

3 years ago

WILL GIVE BRAINLIEST!

andreev551 [17]

Answer:

1,632(

Step-by-step explanation:

I did 34,284/252 to figure out the # of groups with 252 people

I got 136.04-- rounded it to 136

I then multiplied 136*12=1,632

so 1,632 people have type A blood

3 0

3 years ago

Read 2 more answers

What is the length of the base of a right triangle with an area of 20 m and a height of 4 m?

fenix001 [56]

Answer: 5 m

Step-by-step explanation:

height x base = area

4x=20

x=20÷4

x=5

base= 5m

4 0

3 years ago

List the first 11 perfect squares

vlabodo [156]

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121

6 0

4 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

3 years ago