Simplify the expression "the square root of x^10 times the square root of x^4" completely. a. the square root of x^14 b. 2 times the square root of x^10 c. 3 times the square root of x^5 d. x^7

Using the rule that √a times √b = √(a times b)
√(x^10) times √(x^4) = √(x^10 times x^4)
using the rule that x^a times x^b = x^(a+b)
√(x^10 times x^4) = √x^(10 + 4) = √x^14
to cancel out √ with ^14, divide the number 14 by 2 (if it is ∛ then you would divide 14 by 3, and so on, but in this case it is the square root)
√x^14 = x^(14/2) = x^7
so the answer is D. x^7

Let's talk through this a one step at a time.

*Since f(x) is concave-up with its vertex on the x-axis, we know f(x) ≥ 0.

*We also know that when we shift a function's domain by a positive number, we shift the function left and when we shift a function's domain by a negative number, we shift the function right. So f(x-5) is f(x) shifted to the right by 5.

*At this point, f(x-5) has its vertex at (5,0).

*When we negate f(x-5), the parabola becomes concave down yet the vertex remains at (5,0). Now we're at -f(x-5). At this point we have -f(x-5)≤0 with a range (-∞,0]

*If we add 2 to create g(x)=2-f(x-5), then we have a concave down parabola with its vertex shifted up by 2, at (5,2). So, g(x) is concave down with its vertex at (5,2). Hence