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Vinvika [58]
2 years ago
13

Find the missing side

Mathematics
1 answer:
Marrrta [24]2 years ago
5 0

Answer:

x = 71.5 in

Step-by-step explanation:

Remember the SOHCAHTOA rule

Soh...

Sine = Opposite / Hypotenuse

...cah...

Cosine = Adjacent / Hypotenuse

...toa

Tangent = Opposite / Adjacent

In this case, we use: CAH.

cos(52) = 44/x

x = 44/cos(52)

x = 71.5 inches

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You have 59 to spend on lip balm and hand sanitizer. The equation 1.5x + 2.5y = 9 represents this situation, where x is tubes of
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Answer:

Step-by-step explanation:

you have to keep the x and the y sepret and then you can divied one by nine then you get the answer so there you go

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2 years ago
I need x and the two angles
barxatty [35]

Answer:

x= 13

both angles are equal to 106

Step-by-step explanation:

8x + 2= 14x - 76

-8x +76 - 8x +76

78 = 6x

78/6 = 13

8 (13) + 2 = 104 + 2 = 106

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7 0
2 years ago
Find the product simplify the answer 3*1/7
Pani-rosa [81]

Answer:3/7

Step-by-step explanation:

3 * 1/7 is

3 * 1 = 3

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There’s nothing to simply, as that fraction cannot be reduced

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2 years ago
Describe why you can change a subtraction problem into an addition problem
Elena-2011 [213]
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8 0
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

4 0
2 years ago
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