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3 years ago

8

What is the solution of the system?

1 answer:

kakasveta [241]3 years ago

8 0

Divide the first equation by 2 and add the result to the second equation. This will eliminate x.

... (-4x -2y)/2 + (2x +4y) = (-12)/2 +(-12)

... 3y = -18 . . . . . collect terms

... y = -6 . . . . . . . divide by 3

Substitute this into either equation to find x. Let's use the second equation, where the coefficient of x is positive.

... 2x +4(-6) = -12

... 2x = 12 . . . . . . . . add 24

... x = 6 . . . . . . . . . . divide by 2

The solution is (x, y) = (6, -6).

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Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
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$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
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130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

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$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

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$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

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