Let the number be x. The equation is 6x - 4 = 14. Adding 4 gives 6x = 18, so x = 3.

Let the number be x. The equation is 6x + 5 = 41. Subtracting 5 gives 6x = 36, so x = 6.

Let the number be x. The equation is 7x - 14 = 0. Adding 14 to both sides gives 7x = 14. Dividing by 7 gives x = 2.

P(x) = x^2 - 4x + 4 = (x - 2)(x - 2) or (x - 2)^2.

Area = Length × Width = (x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6.

Area = Length × Width = 3x × 2x = 6x^2.

Substituting x = 3 into the function: f(3) = 2(3)^2 - 8(3) + 6 = 18 - 24 + 6 = 0.

Substitute x = 2 into the function: f(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3.

The polynomial can be factored as (x - 2)(x + 2). Setting each factor to zero gives us x - 2 = 0 or x + 2 = 0, so the roots are x = -2 and x = 2.

Substituting x = 2 into the function gives f(2) = 2^2 - 4(2) + 4 = 0.

Factoring gives f(x) = (x - 2)(x - 4). Setting each factor to zero gives the roots x = 2 and x = 4.

Substituting x = 3 into the function: f(3) = 3^2 - 6(3) + 9 = 9 - 18 + 9 = 0.

We can factor the polynomial as f(x) = (x - 3)(x + 3). Setting each factor to zero gives us the roots x = -3 and x = 3.

If one root is 2, then the other root can be found using the product of the roots: 2 * r = 6, so r = 3. The sum of the roots is 2 + 3 = -p, thus p = -5.

Using the sum of the roots, p = -(3 + 1) = -4. The product of the roots gives q = 3 * 1 = 3.

The nth term of a geometric progression is given by a_n = a * r^(n-1). Here, a = 3, r = 2, and n = 4. So, a_4 = 3 * 2^(4-1) = 3 * 2^3 = 3 * 8 = 24.

The nth term of a geometric progression is given by a_n = a * r^(n-1). Here, a = 5, r = 3, and n = 3. So, a_3 = 5 * 3^(3-1) = 5 * 3^2 = 5 * 9 = 45.

The nth term of a geometric sequence is a * r^(n-1). Here, a = 3, r = 2, n = 4. So, 3 * 2^(4-1) = 3 * 8 = 24.

The nth term of an AP is given by a_n = a + (n-1)d. Here, a = 4, d = 5, and n = 10. So, a_10 = 4 + (10-1)5 = 4 + 45 = 49.

Convert to slope-intercept form (y = mx + b).\n1. 3y = -2x + 12\n2. y = -2/3x + 4.\nThe slope is -2/3.

Factoring the quadratic gives (x - 5)(x + 1) = 0. Thus, the roots are x = 5 and x = -1.

Let the other number be y. Then x + y = 15. Therefore, y = 15 - x.

Let the other number be y. Then x + y = 30. Therefore, y = 30 - x.

The polynomial can be factored as (x + 3)(x + 3) = 0, giving the root x = -3.

Factoring gives (x - 2)(x - 2) = 0, so the repeated root is x = 2.

For a double root, the discriminant must be zero: b^2 - 4ac = 0. Here, 4^2 - 4(1)(k) = 0. Thus, 16 - 4k = 0, leading to k = 4.

The polynomial can be factored as (x - 3)(x - 3) = 0, giving a repeated root of x = 3.

Let the number of girls be 2x and boys be 3x. The equation is 2x + 3x = 25. Solving gives 5x = 25, so x = 5. Therefore, girls = 2x = 10.

Divide both sides by 2: x - 3 = 2. Add 3: x = 5.

Distribute: 4x - 4 = 2x + 6. Subtract 2x from both sides: 2x - 4 = 6. Add 4: 2x = 10. Divide by 2: x = 5.