Let's go through the problems from the image step by step.

6) $h(x) = f(x) - g(x)$

Given:

• $f(x) = 3x^2 + 9x - 12$
• $g(x) = 3x^2 - x - 7$

Find $h(x)$ and determine what it's divisible by.

Solution:

$h(x) = f(x) - g(x)$

$h(x) = (3x^2 + 9x - 12) - (3x^2 - x - 7)$

Simplifying:

$h(x) = 3x^2 + 9x - 12 - 3x^2 + x + 7$

$h(x) = (9x + x) + (-12 + 7)$

$h(x) = 10x - 5$

Now, $h(x) = 10x - 5$, and we can factor this:

$h(x) = 5(2x - 1)$

Thus, $h(x)$ is always divisible by 5.

Answer: $\boxed{5}$

7) $X \oplus Y = (X - Y)^2$

We are asked to find $3 \oplus (-2)$.

Solution:

By the definition given, $X \oplus Y = (X - Y)^2$. Substituting the values:

$3 \oplus (-2) = (3 - (-2))^2$

$3 \oplus (-2) = (3 + 2)^2 = 5^2 = 25$

Answer: $\boxed{25}$

8) Sum of 5 consecutive odd integers greater than 20

We need to find the least possible outcome for the sum of 5 consecutive odd integers where all integers are greater than 20.

Solution:

The smallest odd integer greater than 20 is 21. The 5 consecutive odd integers starting from 21 are:

$21, 23, 25, 27, 29$

Now, let's sum them:

$21 + 23 + 25 + 27 + 29 = 125$

Thus, the least possible outcome is 125.

Answer: $\boxed{125}$

9) Rock band concert attendance

• Year 1: Attendance is 25,000.
• Year 2: Attendance is 5% lower than the first year.
• Year 3: Attendance rises by 10% from the second year.

We need to calculate the attendance for the third year.

Solution:

1. For Year 2, we calculate 5% lower than 25,000: $\text{Year 2 Attendance} = 25,000 \times (1 - 0.05) = 25,000 \times 0.95 = 23,750$

2. For Year 3, the attendance increases by 10% from Year 2: $\text{Year 3 Attendance} = 23,750 \times (1 + 0.10) = 23,750 \times 1.10 = 26,125$

Thus, the attendance for the third year is 26,125.

Answer: $\boxed{26,125}$

10) Solve $x^2 + 7x = -10$

We need to solve the quadratic equation $x^2 + 7x + 10 = 0$.

Solution:

The equation can be factored as:

[ x^2 + 7x + 10 = (x +