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3 years ago

Questions 23-26 I need help with

Mathematics

2 answers:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

23. b

24. c

25. b

26. c

Step-by-step explanation:

23. In problem 21, you found the coefficient matrix to be ...

$\left[\begin{array}{ccc}2&-3&1\\1&2&-4\\-3&-2&3\end{array}\right]$

Its determinant is ...

(2)(2)(3) +(-3)(-4)(-3) +(1)(1)(-2) -(-3)(2)(1) -(-2)(-4)(2) -(3)(1)(-3)

= 12 -36 -2 +6 -16 +9 = -27 . . . . choice b

24. Looking at the equation involving x11 terms, we have ...

-3x +7 = 10

-3x = 3 . . . . . subtract 7

x = -1 . . . . . . divide by -3 . . . . matches choices A and C

We note that choices A and C differ in their values for x22, so we consider the equation for that:

-3x -3 = -18

x + 1 = 6 . . . . . . . divide by -3

x = 5 . . . . . . . . matches choice C

25. Dx will be the determinant of the matrix with the constants in the first column. Choice B is the only one that matches.

26. The value of Dx is -78 +24 = -54. The determinant of the coefficient matrix is (5)(6) -(-3)(-4) = 30 -12 = 18. So the value of x is -54/18 = -3, which matches choice C.

suter [353]3 years ago

7 0

See the attached picture:

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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

Andrews [41]

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: