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Alenkinab [10]
3 years ago
11

Multiplies to -2 adds to 6

Mathematics
1 answer:
ikadub [295]3 years ago
6 0
Xy=-2
x+y=6

minnus x from second equation
y=6-x
sub for y
x(6-x)=-2
6x-x^2=-2
add x^2 both sides and minus 6x both sides
0=x^2-6x-2
use quadatic formula

if you have
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}


x^2-6x-2=0
a=1
b=-6
c=-2
x=\frac{-(-6)+/- \sqrt{(-6)^{2}-4(1)(-2)} }{2(1)}
x=\frac{6+/- \sqrt{36+8} }{2}
x=\frac{6+/- \sqrt{44} }{2}
x=\frac{6+/- 2\sqrt{11} }{2}
x=3+/- \sqrt{11}

x=3+ \sqrt{11} or x=3- \sqrt{11}
aprox
x=6.31662 or -0.316625

those are the numbers

6.31662 and -0.316625


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Andreyy89

Answer:

\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}.

Step-by-step explanation:

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In other words,

\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}.

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\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C.

Therefore

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