To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:

\( \frac{6x^8}{4x^3} \)
\( \frac{6}{4} \) x^{(8 - 3)}
1\(\frac{1}{2}\)x⁵

The equation for this sequence is:

a_n = a_n-1 + 4(n - 1)

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 4(6 - 1)
a₆ = 41 + 4(5)
a₆ = 61

The number of students taking German or Spanish is 15 + 5 = 20. Of that group of 20, 3 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 20 - 3 = 17 who are taking at least one language. 23 - 17 = 6 students who are not taking either language.

First, solve for \( \left|y - 1\right| \):

\( \left|y - 1\right| \) - 5 = -3
\( \left|y - 1\right| \) = -3 + 5
\( \left|y - 1\right| \) = 2

The value inside the absolute value brackets can be either positive or negative so (y - 1) must equal + 2 or -2 for \( \left|y - 1\right| \) to equal 2:

So, y = -1 or y = 3.

You have 1 out of the total of 50 raffle tickets sold so you have a (\( \frac{1}{50} \)) x 100 = \( \frac{1 \times 100}{50} \) = \( \frac{100}{50} \) = 3% chance to win the raffle.