Answer: Your answer is r^12t^12
Step-by-step explanation: You use the distribution property.
Answer:
2<x<4/3
Step-by-step explanation:
Given the equation of a graph to be y = |3x− 5|, if the equation is one unit to the right, this can be expressed as |3x-5| > 1.
Solving the resulting equation
|3x-5| > 1.
Since the function 3x-5 is in a modulus sign, this means that the function can take both negative and positive values.
For positive value of the function;
+(3x-5) > 1
3x > 1+5
3x>6
x>6/3
x>2 ... (1)
For the negative value of the function;
-(3x-5) > 1
On expansion
-3x+5 > 1
-3x > 1-5
-3x > -4
Multiplying through by -1 will also change the inequality sign
x < -4/-3
x < 4/3...(2)
Combining equation 1 and 2, we have;
2<x<4/3
y = sqrt(x) .... parent function
y = sqrt(x+2) .... replace x with x+2 to shift 2 units to the left
y = sqrt(x+2)+3 ... add on 3 to move 3 units up
y = sqrt(-x+2)+3 ... replace x with -x to reflect over y axis
<h3>Answer: y = sqrt(-x+2)+3</h3>
Answer: 6
Step-by-step explanation: why the process of elimination is because 9 is sticking like a sore thumb and the last one is too big the first one and the third one are similar but the third one has a square root of 3 so it would be THE FIRST ONE.
