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3 years ago

A triangle has 2 side lenghts 5 and 12. what could the third side length be

Mathematics

2 answers:

NARA [144]3 years ago

8 0

Answer:

Step-by-step explanation:

Square both of them, add them together and square root the answer

Radda [10]3 years ago

8 0

Answer:

7 < x < 17

Step-by-step explanation:

Given 2 sides of a triangle then the third side x is in the range

difference of sides < x < sum of sides , then

12 - 5 < x < 12 + 5 , that is

7 < x < 17

The 3rd side could be 8, 9, 10, 11, 12, 13, 14, 15, 16

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Answer:

Step-by-step explanation:

In 45 45 90 triangles the ratio is a:a:a root 2 so the answer would be 3root 2/2 or c because when multiplied again by root 2 it will get you 3.

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3 years ago

What is 15 tens - 1 gross

vampirchik [111]

Answer:

see the explanation

Step-by-step explanation:

we know that

A gross is equal to 120 ones or ten dozen

what is 15 tens - 1 gross

we know that

15 tens means ----> That you are adding 10, 15 times or multiplying 10 by 15, which gives you

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1 gross means ---> That you are adding 10, 12 times or multiplying 10 by 12

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Convert to word expression

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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:

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

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

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

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

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

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3 years ago