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3 years ago

Multiplies to -2 adds to 6

1 answer:

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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3 years ago

Int(1 \(1 + {e}^{x} )

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Answer:

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

Step-by-step explanation:

The first derivative of the denominator $1 + e^{x}$ is $e^{x}$ . Rewrite the fraction to obtain that expression on the numerator.

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In other words,

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Therefore

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

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