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3 years ago

5

If the average(arithmetic mean) of 5 and r is 7 and the average of 3 and s is 3 , what is the average of r and s?

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer:

C

Step-by-step explanation:

Calculate the values of both r and s, that is

$\frac{5+r}{2}$ = 7 ( multiply both sides by 2 )

5 + r = 14 ( subtract 5 from both sides )

r = 9

and

$\frac{3+s}{2}$ = 3 ( multiply both sides by 2 )

3 + s = 6 ( subtract 3 from both sides )

s = 3

Hence

$\frac{9+3}{2}$ = $\frac{12}{2}$ = 6 → C

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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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