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3 years ago

Please Solve 3x + 6y=27

2 answers:

max2010maxim [7]3 years ago

5 0

You need more information than that to solve for x and y. However, it is possible for you to graph the equation and see what all the possible values for x and y can be. A possible coordinate can be x=3 and y=3.

jek_recluse [69]3 years ago

5 0

I'm not sure what you need to solve for, so I'll solve for both x and y.

3x + 6y = 27
3x = 27 - 6y
x = (27 - 6y)/3
x = 9 - 2y
so there's x
3x + 6y = 27
6y = 27 - 3x
y = (27 - 3x)/6
y = 4.5 - .5x
And there's y

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Answer:

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3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

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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:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

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

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: