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2 years ago

8

Solve the equation. –3x + 1 + 10x = x + 4 x = x = x = 12

1 answer:

Xelga [282]2 years ago

5 0

Remember you can do anything to an equation as long as you do it to both sides

-3x+1+10x=x+4

add like terms
7x+1=x+4
minus x both sides
6x+1=4
minus 1 both sides
6x=3
divide both sides by 6
x=1/2
x=0.5

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What are the roots of the polynomial equation? x2−12x−64=0

Elodia [21]

Answer: " x = 16 ; - 4 " .
_____________________________
<u>Note</u>:
______________________________
-16 , +4 = -12 .

-16 * 4 = -64
______________________________
(x - 16)(x + 4) = 0 ;
______________________________
x = 16 ; - 4 .
_____________________________

8 0

2 years ago

Read 2 more answers

56. How many tangent lines to the curve <img src="https://tex.z-dn.net/?f=y%3Dx%20%2F%28x%2B1%29" id="TexFormula1" title="y=x /(

PIT_PIT [208]

There are 2 tangent lines that pass through the point

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

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Explanation:

Given:

$y=\frac{x}{x+1}$

The point-slope form of the equation of a line tells us that the form of the tangent lines must be:

$y=m(x-1)+2$ $[1]$

For the lines to be tangent to the curve, we must substitute the first derivative of the curve for $m$ :

$\frac{dy}{dx} =\frac{d(x)}{dx}(x+1)-x^\frac{d(x+1)}{dx} \\ \\$

$\frac{dy}{dx} =\frac{x+1-x}{(x+1)^2}$

$\frac{dy}{dx}= \frac{1}{(x+1)^2}$

$m=\frac{1}{(x+1)^2}$ $[2]$

Substitute equation [2] into equation [1]:

$y=\frac{x-1}{(x+1)^2}+2$ $[1.1]$

Because the line must touch the curve, we may substitute $y=\frac{x}{x+1}:$

$\frac{x}{x+1}=\frac{x-1}{(x+1)^2}+2$

Solve for x:

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$x^2+4x+1$

$x\frac{-4±\sqrt{4^2-4(1)(1)} }{2(1)}$

$x=-2$ ± $\sqrt{3}$

$x=-2$ ± $\sqrt{3}$ <em> </em> $and$ $x=-2-\sqrt{3}$

There are 2 tangent lines.

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

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3 years ago

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