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2 years ago

Hey! i’ll give brainliest please help

Mathematics

2 answers:

balu736 [363]2 years ago

8 0

Answer:A!

Step-by-step explanation:

maxonik [38]2 years ago

8 0

Answer:

Step-by-step explanation:

Its A because the plants you see in deserts need water to germinate themselves. Its like cactus. They absorb water like camels store water in their humps.

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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

Jaleel wrote a book. He wants to print some copies for

kotykmax [81]

Yes. there is a fixed cost and variable cost.

8 0

2 years ago

WILL GIVE BRAINLIEST! PLS HELP!!!

velikii [3]

Answer: 140m

Step-by-step explanation:

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2 years ago

Simplify the expression 8−(−2)/7−12

Ymorist [56]

(8−(−2))/7−12

10/-5

-2

3 0

2 years ago

Cos (x + π/ 3) + cos (x - π/ 3) = 1 Solve for "x" on the interval [0, 2pi)

kenny6666 [7]

$\bf cos\left( x+\frac{\pi }{3} \right)+cos\left( x-\frac{\pi }{3} \right)=1\\\\
-----------------------------\\\\\
[cos(x)cos\left( \frac{\pi }{3} \right)\underline{-sin(x)sin\left( \frac{\pi }{3} \right)}] 
+ [cos(x)cos\left( \frac{\pi }{3} \right)\underline{+sin(x)sin\left( \frac{\pi }{3} \right)}]=1
\\\\\\$

$\bf cos(x)cos\left( \frac{\pi }{3} \right)+cos(x)cos\left( \frac{\pi }{3} \right)=1\implies 2cos(x)cos\left( \frac{\pi }{3} \right)=1
\\\\\\
2cos(x)\cdot \cfrac{1}{2}=1\implies cos(x)=1\implies \measuredangle x=cos^{-1}(1)
\\\\\\
\measuredangle x = 
\begin{cases}
0\\
2\pi 
\end{cases}$

8 0

3 years ago