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Murljashka [212]
2 years ago
5

Q : what 1:9 x 5 = ?

Mathematics
2 answers:
mel-nik [20]2 years ago
8 0

Answer:

0.5556 This is the answer

Tema [17]2 years ago
6 0

Answer:

9.5 or 1/45

Step-by-step explanation:

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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

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

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:

x(x+1)=(x-1)+2(x+1)^2

x^2+x=x-1+2x^2+4x+2

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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2 years ago
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