CPhill

Here's 20

Center (1,2)   Point (0,6)   So....we can find the radius using the distance formula

r = √[(1-0)^2  + (6-2)^2] = √[(1) + 4^2] = √[1 + 16] = √17

So...we have

(x-h)^2 + (y-k)^2 = r^2     where (h,k) is the center  and r is the radius

(x-1)^2 + (y-2)^2  = 17

9c + 4 ≤ 22        Subtract 4 from both sides

9c ≤ 18              Divide both sides by 9

c ≤ 2

And that's it......

I actually used WolframAlpha....we could try some synthetic division - plugging in all the possible factors of 21 over all the possible factors of 4 ... this could be a little time consuming. However, I see that 1 is not a root, so I might test -1...so we have....

-1     4   -15        2        21

               -4       19     -21

        4     -19      21        0

So....-1 is a root....and we have the 2nd degree polynomial ......4x^2 -19x +21

Factoring this, we have   (4x -7) (x-3)

So, the factorization is   (x+1)(4x-7)(x-3)

Sorry for not being more thorough the first time, but I was in a hurry  !!!

I assume you want to integrate this with respect to "x"  ??

Simplifying this , first, we have

(y + x^2*y^2) / [(x^2)(y^2)]  =

1/(x^2*y) + 1 =

(1/y)(1/x^2) +1  ........and integrating this, we have

(1/y)∫x^(-2) dx   + ∫ (1) dx =

(1/y) [-x^(-1)] + x  + C  =

x - 1/(xy) + C

When you say "side," I think you mean the "opposite" side.  (It can't be the hypoteneuse because that would have to be longer than 120").

So, the tangent function would relate these. We have:

tan(θ) = 12/120     So...taking the tangent inverse to find "theta", we have

tan^-1(12/120)  ≈ 5.71°

I was only talking about when the deposit was made....at the first of the month is called an "annuity due" - if I remember correctly. Deposits made at the end of the month are known as an "ordinary" annuity. I was assuming the first situation....but  the second could be assumed also...... it's not stated in the problem....

Ah.....I see my error here....we're picking cards one at a time....!!!

OK...I assumed "dealt"

Sorry!!

Thanks, Bertie...I see where I made my mistake !!!!

Good one , Anon.    !!!!

(4i)^2 =  16i^2   and i^2 = -1 .... so we have .....  16(-1) = -16