proyaop

When you divide by a fraction, you multiply by the reciprocoal, which is 1/(10/11) = 11/10 (flip numerator with denominator).

Thus, 5/9 divided by 10/11 is 5/9 times 11/10 = 55/90. Divide top and bottom by 5 and you get 11/18.

Thanks melody, I liked your reasoning.

Yes the answer was k = 1 for the optimal solution --- thanks, proyaop.

There is no greatest common factor.

First, get 0 on one side.

x^2 + 16x - 22 = 0

Factors of 22: (22, 1) and (11, 2)

We see that there is no factors of 22 that add/subtract to get 16.

Complete the square:

x^2 + 16x + 64 = 86

(x + 8)^2 = 86

$x + 8 = \pm\sqrt{86}$

$x = -8 \pm \sqrt{86}$

$9^y = {3*6*27*4\over81*2}$

$9^y = {81 * 6 * 2 * 2\over81 * 2}$, cancel terms (81 and 2)

$9^y = {1 * 6 * 2 * 1\over1*1}$

$9^y = 12$

y is approximately 1.131

Combine like terms:

x^2 + y^2 + 20x - 8y - 48 = 0

(x + 10)^2 + (y - 4)^2 - 164 = 0

radius is square root of 164.

Area = $2\pi\sqrt{41}$

$2x-7\over\sqrt{x^2-7x+6}$

That means the denominator has to be Greater or Equal to 0.

(x-6)(x-1) = 0, x = 1 and 6 for values of 0. 

x^2 - 7x + 6 > 0 

(x - 6)(x - 1)>0, double neg or double pos.

x>6 and x<1

Domain: (-inf, 1] U [6, inf)

1/3 < M/30<5/3

Multiply by 30

10

M is from 11 - 49 inclusive, 49 - 11 + 1 = 39 integer values of M

The x coordinate can never reach the value of 3/2, but the y value can.

After some calculations, you'll see that when t = 25/18, it reaches 3/2. Then add 2 every time, so the second time it reaches will be at 61/18, then 97/18 then 133/18. 

${25\over18}, {61\over18}, {97\over18}, {133\over18}$

By the angle bisector theorem, BC/AB = CY/AY

BC/8 = 3/8

BC = 3

Notice how a long of the terms share a denominator of a^3 + 7, so we can multiply every term in the equation by a^3 + 7.

1 - 7(a^3 + 7) = -a^3 + 5(a^3 + 7)

a^3 + 1 = 12(a^3 + 7)

a^3 + 1 = 12a^3 + 84

11a^3 = -83

a = cube root of -83/11