1.What is the domain of the function?
Y=√3x+3
A.x≤-1
B.x>1
C.x≥-1
D.x<1
I think the answer is B please correct me if I'm wrong
I will assume from looking at the answers that you meant
y = √(3x+3)
Since we only take √'s of non-negative numbers,
3x+3 ≥ 0
3x ≥ -3
x ≥ -1
Can you check 3 more for me please
1.Simplify the radical expression.
√5+6√√5
A.5√5
B.7√10
C.7√5
D.5√10
2.Simplify the radical expression.
2√6+3√96
A.14√6
B.14√96
C.5√96
D.50√6
3.Simplify the radical expression.
(8+√11)(8-√11)
A.53
B.75+16√11
C.-57
D.64+√11
1.C
2.A
3.A
To determine the domain of the function y = √(3x + 3), we need to find the values of x that are allowed.
The square root of a number is only defined if the number inside the square root is non-negative, so we set the expression 3x + 3 greater than or equal to zero:
3x + 3 ≥ 0
Next, we solve for x:
3x ≥ -3
Dividing both sides by 3 (since 3 is positive):
x ≥ -1
Therefore, the domain of the function is all real numbers greater than or equal to -1.
Option C.x≥-1 is the correct answer. Good job!