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artcher [175]
3 years ago
15

Please help asap!!!!!! (picture)

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
6 0

Answer:6-\sqrt{21} , 6+\sqrt{21}

Step-by-step explanation:

For x^{2} =-12x+15, so x^{2} +12x-15=0 We apply the quadratic equation formula:(-b+-\sqrt{b^{2}-4*a*c })/2*a, where a=1, b=12 y c=-15. so;

(12+-\sqrt{12^{2} -4*1*(-15)} )/(2*1) = (12+-\sqrt{144-60})/2 = (12+-\sqrt{84})/2, but \sqrt{84} = 2\sqrt{21} We have that:

Solution : (6-\sqrt{21} , 6+\sqrt{21} )

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Integrate this two questions ​
tankabanditka [31]

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\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)

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5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

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using the same approach as above.

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Answer:

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