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Vladimir79 [104]
3 years ago
7

Simplify the expressions. show your work. 3x(4x4 – 5x)

Mathematics
2 answers:
vovangra [49]3 years ago
5 0
3x(4x^4-5x)=(3x)(4x^4)-(3x)(5x)=(3\cdot4)x^{1+4}-(3\cdot5)x^{1+1}\\\\=\boxed{12x^5-15x^2}
hammer [34]3 years ago
3 0
3x(4x4 – 5x)
= 12x^5-15x^2
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Timmy has 13 dollars in his bank account he withdrew m amount of dollars. He is now 100 dallors in debt. How much
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$113

Step-by-step explanation:

Since he had $13 and now he is $100 in debt:

First, he withdrew $13 to get $0 in his bank account.

He then withdrew $100 to get -$100 in his bank account.

13 + 100 = 113

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Which table represents a linear function?
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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.

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.

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