They left $6.72 for a tip.
42 x 0.16 = 6.72
Answer:
Here we have the domain:
D = 0 < x < 1
And we want to find the range in that domain for:
1) y = f(x) = x
First, if the function is only increasing in the domain (like in this case) the minimum value in the range will match with the minimum in the domain (and the same for the maximums)
f(0) = 0 is the minimum in the range.
f(1) = 1 is the maximum in the range.
The range is:
0 < y < 1.
2) y = f(x) = 1/x.
In this case the function is strictly decreasing in the domain, then the minimum in the domain coincides with the maximum in the range, and the maximum in the domain coincides with the minimum in the range.
f(0) = 1/0 ---> ∞
f(1) = 1/1
Then the range is:
1 < x.
Notice that we do not have an upper bound.
3) y = f(x) = x^2
This function is strictly increasing, then:
f(0) = 0^2 = 0
f(1) = 1^2 = 1
the range is:
0 < y < 1
4) y = f(x) = x^3
This function is strictly increasing in the interval, then:
f(0) = 0^3 = 0
f(1) = 1^3 = 1
the range is:
0 < y < 1.
5) y = f(x) = √x
This function is well defined in the positive reals, and is strictly increasing in our domain, then:
f(0) = √0 = 0
f(1) = √1 =1
The range is:
0 < y < 1
Answer:
x=2
Step-by-step explanation:
5(x-5)=3(x-7)
Distribute
5x-25 = 3x-21
subtract 3x from each side
5x-25-3x = 3x-21-3x
2x-25 = -21
Add 25 to each side
2x-25+25 = -21 +25
2x=4
Divide by 2
2x/2 =4/2
x=2
Answer:
d
Step-by-step explanation:
The rule (x, y ) → (x + 4, y - 6 )
means add 4 to the original x- coordinate and subtract 6 from the original y- coordinate, thus
P(- 8, 3 ) → P'(- 8 + 4, 3 - 6 ) → P'(- 4, - 3 )
Q(- 8, 6 ) → Q'(- 8 + 4, 6 - 6 ) → Q'(- 4, 0 )
R(- 3, 6 ) → R'(- 3 + 4, 6 - 6 ) → R'(1, 0 )
This is an infinite loop display
