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3 years ago

Solve the quadratic equation. (x - 5)2 = 9

1 answer:

5 0

(x - 5)^2 = 9
(x - 5)(x - 5) = 9
x^2 - 10x + 25 = 9
x^2 - 10x + 25 - 9 = 0
x^2 - 10x + 16 = 0
(x - 8)(x - 2) = 0

x - 8 = 0
x = 8

x - 2 = 0
x = 2

solution : x = 8 and x = 2

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Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

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

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

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The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

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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:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

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3 years ago

Read 2 more answers

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3 years ago

Write the slope-intercept form of the equation of the line; (-5,-4) and prep. To y equals 5x plus 1

ohaa [14]

Answer:

$y=-\frac{1}{5} x-5$

Step-by-step explanation:

The given line is defined by: $y=5x+1$ , where we see that the slope is 5 and the y-intercept 1.

In order to find a line perpendicular to the given one, we need it to have a slope that is the "opposite of the reciprocal" of the given slope.

"Opposite" means it would have its sign inverted (in our case from positive to negative); and "reciprocal means that instead of 5, it would be its reciprocal: $\frac{1}{5}$ .

We can write this new line with such slope, and try to find its y-intercept (b) by using the given condition that requires it to go through the point (-5,-4) on he plane:

$y=-\frac{1}{5} x+b$

we require then that when $x=-5$ , the value of $y=-4$ .

Therefore: $-4=-\frac{1}{5} (-5)+b\\-4=\frac{5}{5} +b\\-4=1+b\\b=-4-1=-5$

Then our final answer is that the new line should have the form: $y=-\frac{1}{5} x-5$

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