Find distance between -5 and 17

Mathematics

1 answer:

lana [24]3 years ago

4 0

The distance between the two is 22 units.

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consider the circle x^2+y^2-6x-8y=0. a.) find an equation of the tangent line to the circle at the point (0,0). b.) find the cen

inna [77]

The slope of the tangent line of the circle $x^2+y^2-6x-8y=0$ is $\frac{dy}{dx}$ :

to find it we use implicit differentiation:

$\frac{dy}{dx}(x^2)+\frac{dy}{dx}(y^2)-\frac{dy}{dx}(6x)-\frac{dy}{dx}(8y)=0\\\\2x+2y \cdot\frac{dy}{dx}-6-8\frac{dy}{dx}=0\\\\\frac{dy}{dx}(2y-8)=-2x+6\\\\\frac{dy}{dx}= \frac{-2x+6}{2y-8}= \frac{-x+3}{y-4}$

thus the slope of the tangent line at a point (x, y) of the circle is:

$m= \frac{dy}{dx}=\frac{-x+3}{y-4}$

part a:

m at (0, 0) is (-0+3)/(0-4)=3/(-4)=-3/4

the equation of the tangent line is

(y-0)=(-3/4)(x-0)

y=(-3/4)x

part b)

The equation of the circle can be written in standard form by completing the square:

$x^2+y^2-6x-8y=0\\\\x^2-6x+y^2-8y=0\\\\(x^2-6x+9)-9+(y^2-8y+16)-16=0\\\\(x-3)^2+(y-4)^2=5^2$

thus the circle has radius (3, 4) and radius 5.

part c.

$m=\frac{-x+3}{y-4}=\frac{-6+3}{0-4}= \frac{-3}{-4}= \frac{3}{4}$

the equation of the line is:

y-0=(3/4)(x-6)

y=(3/4)x-9/2

d) the lines are y=(-3/4)x and y=(3/4)x-9/2

they meet at x:

(-3/4)x=(3/4)x-9/2

(-6/4)x=-9/2

(6/4)x=9/2

(2/2)x=3/1

x=3,

at x=3, y=(-3/4)x=(-3/4)*3=-9/4

Check the graph, generated using desmos.com

4 0

3 years ago

MULTIPLE CHOICE

cricket20 [7]

Add the equations together so that you get 6x=-4, how I got their is cancel out the +/-3 and then add the -10+6=-4. Then you’d bring over the 6 and it would look like this x=-4/6 which you would simplify by dividing 2 into the 4 and 6 and should look like this x=-2/3

3 0

4 years ago

PLEASE HELP ME NOW!!!!!!!!!!!!!!!!!!!!!!!!

Natali5045456 [20]

Answer

0 is the answer

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Solve for<br> 5(x - 10) = 3015<br> x 8

sveta [45]