BuilderBoi

Simplify the left-hand side: \({3^2}^y = 3^{2y}\). 

Note that the exponents have the same bases, so we have \(2y = 6\). 

Now, just solve for y. 

The common ratio of the series is \(b \over 7\), and the starting term is \(b \over 7\), so we have the equation: \(\large{{{b \over 7} \over 1 - {b \over 7}} = 7}\)

Multiplying the entire equation by \(1 - {b \over 7}\) gives us \({b \over 7} = 7(1 - {b \over 7})\), which simplifies to \({b \over 7} = 7 - b\). 

Multiplying the entire equation by 7 gives us \(b = 49 - 7b\), meaning \(b = \color{brown}\boxed{49 \over 8}\)

Let the total amount of students in the class be \(s\). 

Note that \({5 \over 9}\)th of the students will get 4 sweets, \({1 \over 3} \times (1 -{5 \over 9}) = {4 \over 27}\)th of the students will get 3 sweets, and \({8 \over 27}\)th of the students will get 2 sweets. 

From this, we can form the equation: \(({5 \over 9}s \times 4) + ({4 \over 27}s \times 3) + ({8 \over 27}s \times 2) = 264\). 

Simplifying, we get: \({88 \over 27}s = 264\), meaning there were originally \(\color{brown}\boxed{81}\) students. 

What about 5,5?

Recall the formula for probability: \(\text{succsess} \over \text{total outcomes}\).

In this case, the successes are when it is caffeine-free and fruity, and the total outcomes are the sum of all the outcomes.

Can you take it from here? 

406) We know that \(x^2 - 6x \) equals either 9 or -9. Solve for x in both cases, and plug them back in to make sure they work. 

405) First, we need to find x, so we have the equation: \(4x + 3 = -9\). Solve for x, and substitute this value into the function. 

404) We need to transform x, which we can do by substituting \(x = x - 1\). This way, it becomes \(f((x-1) + 1) = f(x)\)

So, it would now become \((x-1)^2 + 3(x-1) + 5\).