CPhill

1.7724538509055159

We used to say "All Students Take Calculus"...but I like "All Stations To Central," too!!  (It sounds like something right out of the old TV program..."Car 54, Where Are You??" )

I could have made an error - or even misread the question...let's go through this again.....

m arc AEB =

360 - m arc AB =

360 - 84 =

276

Now that I look at it, I think I did goof on this one before.......please accept my apologies....I try to do a good job on here, but every now and then, I b**w a fuse.........!!!!

Sasini did the binomial expansion of (p - 3)^3

I think what you might have asked for is to either factor, or find the roots of, p^3 - 27...I'll do both....

Factoring is just the difference of two cubes...so we have

(p - 3) * (p^2 + 3p + 9)

We can set these factors equal to 0 to find the roots......setting (p - 3) = 0  gives us that p = 3.

The second factor - (p^2 + 3p + 9) - set to 0, has no real solutions (roots).

I think that's it......

Here's a pretty good page that will help you with this.....

What did you add??

It's in the 4th quadrant because the angle "x" lies between -pi/2  and 0.....

And remember that in the 4th quadrant, x is positive and y is negative. And the tanx = y/x = -12/5

Does that make sense, xvxvxv ???

This is just a Pythagorean right triangle in the 4th quad where x = 5, y = -12, and the hypoteneuse (r) = 13.

And the cosine is given by....

x/r  = 5/13  = Answer (c)

Note that we can write 8^200 as (8^2)^100

And we can write  5^300 as (5^3)^100   .... so we have......

(8^2)^100   ???  (5^3)^100      ....and we can simplify this as......

(64)^100   ???  (125)^100

And it's obvious that ......

(64)^100  <  (125)^100

So....it's true

8^200 < 5^300  !!!!!!

Very nice, Melody....that is well laid out !!!!

3 points from me!!!