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
Answer:
Step-by-step explanation:
The way I did this was I noticed that the graph had a y-intercept of 4 and the only exponential function that had a y-intercept of 4 was
A)Plugging in our initial statement values of y = 16 when x = 10, we get:
16 = 10k
Divide each side by 10 to solve for k:
16/10=
k = 1.6
Solve the second part of the variation equation:
Because we have found our relationship constant k = 1.6, we form our new variation equation:
y = 1.6x
Since we were given that x, we have
y = 1.6()
y = 0
B)Plugging in our initial statement values of y = 1 when x = 15, we get:
1 = 15k
Divide each side by 15 to solve for k:
1/15
=15k
k = 0.066666666666667
She completed 3/4 of her homework
