will mark brainliest!!

3x² * (6x² + 4x - 8) = 18x⁴ + 12x³ - 24x²

2x * (9x³ + 6x² - 12x) = 18x⁴ + 12x³ - 24x²

x² * (18x² + 12x - 24) = 18x⁴ + 12x³ - 24x²

6x² * (3x² + 2x - 4) = 18x⁴ + 12x³ - 24x²

Step-by-step explanatation:

Let's set unknown variables as y (and y₁, y₂, etc)

Question 1:

y *  (6x² + 4x - 8) = 18x⁴ + 12x³ - 24x²

So, divide both sides by (6x² + 4x - 8)

y = 3x²

Question 2:

2x * (9x^y₁ + y₂*x^y₃ - y₄*x) = 18x⁴ + 12x³ - 24x² '

Divide both sides by 2x.

(9x^y₁ + y₂*x^y₃ - y₄*x) = 9x³ + 6x² -12x

y₁ = 3, y₂ = 6, y₃ = 2, y₄ = 12

Question 3:

x^y * (18x² + 12x - 24) = 18x⁴ + 12x³ - 24x²

Divide both sides by (18x² + 12x - 24)

x^y = x²

Question 4:

y₁  * x^y₂ * (3x² + 2x - 4) = 18x⁴ + 12x³ - 24x²

Divide both sides by (3x² + 2x - 4)

y₁  * x^y₂ = 6x²

y₁ = 6, y₂ = 2

I hope this helps! Feel free to ask any questions!

Not enough information - are there 18 yellow balls?

If so, you are trying to calculate the probability at least one of the balls is yellow,

P(no balls are yellow)=P(first ball is non-yellow)×P(second ball is non-yellow)  

P(no yellow)=1230×1129  

P(at least one yellow)=1−P(no yellow)  

You can also calculate the probability of one ball being yellow and them both being yellow:

P(both)=1830×1730  

P(one)=P(first is yellow and second is not)+P(second is yellow and first is not)=1830×1229+1230×1829  

which is just  2×P(first is yellow and second is not).  

Honestly though, this looks just like a homework problem

Step-by-step explanation: