Consider the equation below. -2(bx-5)=16 Answer quick

Mathematics

1 answer:

Vlada [557]3 years ago

6 0

Answer:

b= −3 /x

Step-by-step explanation:

Step 1: Add -10 to both sides.

−2bx+10+−10=16+−10

−2bx=6

Step 2: Divide both sides by -2x.

−2bx

−2x

b= −3 /x

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The curve y = |x|/ 5 − x2 is called a bullet-nose curve. find an equation of the tangent line to this curve at the point (2, 2).

diamong [38]

We have the following curve:

$y=\frac{\left | x \right |}{5-x^2}$

So we need to find an equation of the tangent line to this curve at the point $(2,2)$ . So let's find out if this point, in fact, belongs to the curve:

 $If \ x=2 \\ \\ y=\frac{\left | 2 \right |}{5-2^2}=\frac{\left | 2 \right |}{5-4}=2$ .

We also know that:

 $\left | x \right |= \left \{ {{x \ if \ x \ \geq 0} \atop {-x \ if \ x\ \textless \ 0}} \right.$ 

Given that the point is:

 $(2,2)\ that \ is: \\ x=2\ \textgreater \ 0$

Then we will say that:

$\left | x \right |=x$

Therefore:

$y=\frac{x}{5-x^2}$

Computing the derivative:

$y=\frac{x}{5-x^2} \\ \frac{dy}{dx}= \frac{(1)(5-x^2)-(-2x)x}{(5-x^2)^2}=\frac{5+x^2}{(5-x^2)^2}$

So the derivative solved for $x=2$ is in fact the slope of the line at the point $(2,2)$ , then:

$\frac{dy}{dx} |_{x=2}=\frac{5+x^2}{(5-x^2)^2}|_{x=2}=\frac{5+2^2}{(5-2^2)^2}=9 \\ \\ \therefore m=\frac{dy}{dx} |_{x=2}=9$

Finally, the tangent line is:

$y-y_1=m(x-x_1) \\ \therefore y-2=9(x-2) \\ \therefore \boxed{y=9x-16}$

This is shown in the figure below.