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3 years ago

Please help me out with this worksheet

Mathematics

1 answer:

Sindrei [870]3 years ago

3 0

1.5-6-12n. 1-12n
2.-4+40x-35. -39+40x
3.-36x+6+6. 12-36x
4.40n-32+8n. 48n-32
5.7n-28-6. 34-7n
6.-16+20x+3x 23x-16
7. -2n-16+8. -2n-8
8.25b-25-1. 25b-26
9.-4n+15-6n. -10n+15
-1-3x+6. -3x+5

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Evaluate the following expression:<br> -X- 6x + 10 when x = 5

BigorU [14]

Answer:

-25

Step-by-step explanation:

Evaluate -x - 6 x + 10 where x = 5:

-x - 6 x + 10 = -5 - 6×5 + 10

-6×5 = -30:

-5 + -30 + 10

-5 - 30 + 10 = 10 - (5 + 30):

10 - (5 + 30)

5 + 30 = 35:

10 - 35

10 - 35 = -(35 - 10):

-(35 - 10)

| 3 | 5

- | 1 | 0

| 2 | 5:

Answer: -25

8 0

2 years ago

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: