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Masja [62]
2 years ago
12

Rina draws a plan of her school on a coordinate grid. She plots the school gate at point p(-8,-1) and the library at point Q(2,5

). She notices that the science lab is also located on PQ and is two thirds the distance from the library to the school gate. What are the coordinates of the science lab ?
Mathematics
1 answer:
natka813 [3]2 years ago
5 0

Answer:

The coordinates of the science lab is:

x = -14/3  ,  y = 1

Step-by-step explanation:

∵ P is (-8 , -1) and Q is (2 , 5)

∵ The point of the science lab is (x , y)

∵ The science lab point divided PQ at ratio 2 : 1 from Q (two thirds PQ)

∴ x=\frac{m_{1}x_{p}+m_{2}x_{q}}{m_{1}+m_{2}}

∴ y=\frac{m_{1}y_{p}+m_{2}y_{q}}{m_{1}+m_{2}}

∴ x=\frac{2(-8)+1(2)}{1+2}=\frac{-14}{3}

∴ y=\frac{2(-1)+1(5)}{2+1}=\frac{3}{3}=1

∴ The point of science lab = (-14/3 , 1)

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Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

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Where:

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t - Time, measured in days.

\tau - Time constant, measured in days.

The solution of the differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}

\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }

The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2

The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

The approximate difference in the half-lives of the isotopes is 66 days.

4 0
3 years ago
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Please answer quickly Ill give brainliest.
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Answer:

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Step-by-step explanation:

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Hopefully I was helpful!

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