CPhill

$$\frac{3}{5}$$

Who says an old dog can't learn new tricks????

Where's my points ??.......Where's my points ??.....Where's my points ??..........Where's my points ??..........Where's my points ??.........

DS...I don't mean to be critical, but you're answering a question that was already answered correctly many moons ago....and you gave the wrong answer....!!!

C'mon, dude!!.......we can't give you any points for that......

[72/(-4)+3(-4)] / [5-(-5)]  =

[  -18 - 12 ] / [ 10 ] =

[ -30 ] /  10 ] =

-3

arc AE = [360 - arc AB - arc BC - arc CD - arc DE] =

              [360 -     84   -    38   -    64    -   60  ] =  114

Here's the graph....the red line is y = -x - 7, and the blue line is y = 2x + 5

Multiply both sides of the equation by 4

12x - y = 28  Add y to both sides

12x = y + 28

y + 28 = 12x     and we can write this as

[y - (-28)] = 12(x - 0)

And that's in point-slope form

You already have the point-slope form...I think you might mean slope-intercept form...

Add 2 to both sides

y = 2(x + 3) + 2    simplify

y = 2x + 6 + 2

y = 2x + 8

And that's the slope-intercept form.

It depends upon what "n" might be.......as is.....all we can say is that it equals 2n

The diameter is twice the radius.....however.....if you're asking how to construct a diameter in a circle, here's one way using Euclid's simple method for finding the center of a circle.....the line CE that bisects the chord AB at right angles is a diameter........

You can do both of these with the on-site calculator......