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
2 (n+1) + n² + 4 (n-3)
= 2n + 2 + n² + 4n - 12 ...expand by using distributive property
= n² + 6n - 10 ...combine like terms
hope it helps
Answer:
81
Step-by-step explanation:
Answer:
The answer on A P E X is bal(132)
