<span>1.Describe how the graph of y = x2 can be transformed to the graph of the given equation.
y = (x+17)2
Shift the graph of y = x2 left 17 units.
2.Describe how the graph of y= x2 can be transformed to the graph of the given equation.
y = (x-4)2-8
Shift the graph of y = x2 right 4 units and then down 8 units.
.Describe how to transform the graph of f into the graph of g.
f(x) = x2 and g(x) = -(-x)2
Reflect the graph of f across the y-axis and then reflect across the x-axis.
Question 4 (Multiple Choice Worth 2 points)
Describe how the graph of y= x2 can be transformed to the graph of the given equation.
y = x2 + 8
Shift the graph of y = x2 up 8 units.
Question 5 (Essay Worth 2 points)
Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch.
f as a function of x is equal to the square root of x and g as a function of x is equal to 8 times the square root of x
f(x) = √x, g(x) = 8√x
vertical stretch factor 8
Plz mark as brainlest</span>
Answer:
5/11
Step-by-step explanation:
Black= 22 b+17 g
Brown= 4 b +18 g
Blonde= 25 b+ 10 g
Red= 21 b+ 15 g
Total number= 22+17+4+18+25+10+21+15= 132
Number of girls= 17+18+10+15= 60
Girls are 60/132= 2 × 2 × 3 × 5/ 2 × 2 × 3 × 11= 5/11 of all students
Answer:
3.1782437e+42
Step-by-step explanation:
Answer:
11 am
Step-by-step explanation:
Bus A and Bus B leave the bus depot at 9 am.
Bus A takes 30 minutes to complete its route once
Bus B takes 40 minutes to complete its route once.
We solve this finding the Lowest Common Multiple of the minutes each bus uses to complete it's route
30 = 3 × 10
40 = 4 × 10
= 3 × 4 × 10
= 120 minutes
120 minutes after 9 am is
60 minutes = 1 hour
60 minutes = 1 hour
= 2 hours.
9am + 2 hours
= 11 am.
Therefore, they be back at the bus depot together at 11 am
Average rate of change can be calculated by determining the
rate of change at x = a, and at x = b
f’(x) =2 (3^x) ln(3)
f’(0) = 2 ln(3)
f’(1) = 6 ln(3)
f’(2) = 18 ln(3)
f’(3) = 54 ln(3)
Average:
at section A = [6 ln(3) – 2 ln(3)]/1 = 4 ln(3)
at section B = [54 ln(3) – 18 ln(3)]/1 = 36 ln(3)
section B is 9 times larger.
Based from the f’(x), f’(x) varies as the power of x. so the
greater of value of x, the greater the rate of change.
