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zzz [600]
3 years ago
15

What are the possible steps involved in solving the equation shown? Select three options. 3.5 + 1.2(6.3 – 7x) = 9.38 Add 3.5 and

1.2. Distribute 1.2 to 6.3 and –7x . Combine 6.3 and –7x . Combine 3.5 and 7.56. Subtract 11.06 from both sides. Mark this and return Save and Exit Next Submit
Mathematics
2 answers:
motikmotik3 years ago
8 0

Answer:

1) Distribute 1.2 to 6.3 and -7x

2)Combine 3.5 and 7.56

3)Subtract 11.06 from both sides

Step-by-step explanation:

3.5 + 1.2(6.3 - 7x) = 9.38

Distribute 1.2 to 6.3 and -7x

3.5 + 1.2* 6.3 - 1.2 * 7x = 9.38

        3.5 + 7.56 - 8.4x = 9.38

Combine 3.5 and 7.56

11.06 - 8.4x = 9.38

Subtract 11.06 from both sides

11.06 - 8.4x -11.06 = 9.38 - 11.06

                    -8.4x = -1.68

To find solution:

Divide both sides by (-8.4)

-8.4x/-8.4 = -1.68/-8.4

 x = 0.02

VLD [36.1K]3 years ago
8 0

Answer:

Distribute 1.2 to 6.3 and -7x

Combine 3.5 and 7.56

Subtract 11.06 from both sides

Step-by-step explanation:

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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:

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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}

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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

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