answer:To simplify (sqrt{12}), you factorize the number under the square root to make simplification easier, looking for square factors because they can be taken out of the square root. The number 12 can be factorized into (4 times 3), where 4 is a perfect square. So, [ sqrt{12} = sqrt{4 times 3} = sqrt{4}sqrt{3} = 2sqrt{3} ] Therefore, the simplified form of (sqrt{12}) is: a. (2sqrt{3})

answer:To solve the equation (log_{3}(x-6) + log_{3}(5) = 2log_{3}(2)), let's use the properties of logarithms to simplify and solve for (x). 1. **Combine the logarithms on the left side**: The sum of logarithms with the same base can be rewritten as the logarithm of the product of their arguments, based on the property (log_b(m) + log_b(n) = log_b(mn)). [ log_{3}(x-6) + log_{3}(5) = log_{3}((x-6)5) ] 2. **Use the power property on the right side**: The equation (2log_{3}(2)) can be rewritten using the power rule (nlog_b(m) = log_b(m^n)), which in this case means: [ 2log_{3}(2) = log_{3}(2^2) = log_{3}(4) ] 3. **Set the arguments equal to each other**: Since the bases of the logarithms are the same and their summed/transformed values are equal, we can equate the arguments of the logarithms: [ (x-6)5 = 4 ] 4. **Simplify and solve for (x)**: [ 5(x - 6) = 4 ] [ x - 6 = frac{4}{5} ] [ x = frac{4}{5} + 6 ] [ x = frac{4 + 30}{5} = frac{34}{5} ] Therefore, the solution for (x) is: b. (frac{34}{5})

answer:To solve the inequality (3(x + 8) < x + 4), let's first distribute the (3) on the left side and then collect like terms. [3(x + 8) < x + 4] [3x + 24 < x + 4] Subtract (x) from both sides: [2x + 24 < 4] Subtract (24) from both sides: [2x < -20] Divide both sides by (2): [x < -10] Thus, the solution to the inequality is: a. (x < -10)

answer:Given that (f(x)) is a cubic polynomial that satisfies (f(0) = f(1) = f(2) = 0), it can be described by the cubic polynomial form of (f(x) = k(x)(x-1)(x-2)), where (k) is a constant. This is because a cubic polynomial with three distinct roots, (x = 0), (x = 1), and (x = 2), must factor through those roots. Similarly, given (g(x)) is a quadratic polynomial satisfying (g(0) = g(1) = 0) and (g(2) = 2), it can be described by the form (g(x) = m(x)(x-1)), where (m) is a constant. Given that (g(2) = 2), we can find (m) by evaluating: [g(2) = m(2)(2-1) = 2m = 2] [m = 1] So, (g(x) = x(x-1)). To find (f(3)), we know (f(3) = 6). [f(3) = k(3)(2)(1) = 6k] Since (f(3) = 6), then: [6k = 6] [k = 1] Therefore, (f(x) = x(x-1)(x-2)). Now, we can find (f(-1)): [f(-1) = (-1)(-1-1)(-1-2) = (-1)(-2)(-3) = -6] And (g(-1)): [g(-1) = (-1)(-1-1) = (-1)(-2) = 2] The sum (f(-1) + g(-1)) is: [-6 + 2 = -4] So, the numerical value of (f(-1) + g(-1)) is: a. -4