Please help tysm! Why is the domain of f(x) = cube root of x NOT restricted while the domain of g(x) = sqrt(x) is?
1. It is easier to take the square root of a number than it is to take the cube root of a number.
2. We cannot take the square root of any numbers except for those that are perfect squares, such as 0, 1, 4, 9, 16, etc.
3. This is not true. The domains of both functions are restricted since the cube root and square root of negative numbers are both not real.
4. The cube root of a negative number is still a real number, but the square root of a negative number is not real.

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation:

answer: get a conversion sheet out i dont know the exact measurments

step-by-step explanation: divide the number of feet with the nu,ber of milrs and there your answer

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.