CPhill

For the first one, let the points A =(0, -2) and B = (0, 4) be endpoints of a diameter of a circle with a radius of 3 centered at (0,1).

Then,  the locus of points formed when QA is perpendicular to QB will lie on this circle and it will have the equation x^2 + (y - 1)^2 = 9. Note that, except at the endoints of the diameter, QA and QB will always from two legs of a right triangle with the diameter AB as a hypotenuse.  {To be technically correct, the points A and B shouldn't actually be included in this locus, since the diameter can't be perpendicular to itself.}

Here's a pic

For the second, this is just a parabola with a vertex of (0, 7/2) and a width of "a"

The equation is y = ax^2 + (7/2).....here's the graph when a = 1......

1 mile = 5280 ft....so...

230 / 5280  =

23 / 528  = .04356 mile

It depends upon what you are evaluating......

Sin  = opp / hyp

but.......

Csc = hyp / opp

I'm answering this.....  5√(20) / 49 =

[5 √(4 * 5)] / 49  =

[5 * √4 * √5] / 49 = 

[5*2 * √5 ] / 49 =

[10√5]  / 49 =

(10/49)√5

Whether this is what was asked.....I don't know......

 4pq^2-2p^2q

Take out the greatest common factor = 2pq   .....and we have

2pq (2q - p)

And that's it.....!!!

y = 2x

Let x = 0  , then y = 0

Then, let x = 1, so, y = 2

So....we have the points  (0, 0) and (1, 2)

Now......just plot these on a graph and draw a line through them.....that will be your graph.....

√(1/9) =

√1 / √9  =

1 / 3

We have  (3 x 11)  + (4 x 6) + (4 x 2)  =

33 + 24 + 8  =  65cm^2

Good job, rosala.....

What rosala is saying......in not so many words...is that the nth term is just 3ⁿ

The total number of possible sets that can be made from choosing 9 cards from 52 =

C(52,9)  = 3,679,075,400

But....the ones we are interested in have us choosing any 4 clubs from 13, and from the other 39 cards, selecting any 5 of them.

So we have

C(13.4) * C(39,5)  = 411,666,255

So

411,666,255 / 3,679,075,400  = about 11.2%