Answer:
Yes, an arrow can be drawn from 10.3 so the relation is a function.
Step-by-step explanation:
Assuming the diagram on the left is the domain(the inputs) and the diagram on the right is the range(the outputs), yes, an arrow can be drawn from 10.3 and the relation will be a function.
The only time something isn't a function is if two different outputs had the same input. However, it's okay for two different inputs to have the same output.
In this problem, 10.3 is an input. If you drew an arrow from 10.3 to one of the values on the right, 10.3 would end up sharing an output with another input. This is allowed, and the relation would be classified as a function.
However, if you drew multiple arrows from 10.3 to different values on the right, then the relation would no longer be a function because 10.3, a single input, would have multiple outputs.
Answer:
$21.33
Step-by-step explanation:
If the waiter earns $128 for 6 hours, divide 128 by 6.
21.33...
Answer:
y =4(x-3)^2+2 => y=4x^2-24x+38
y=2(x-3)^2+4 => y=2x^2-12x+22
y=2(x+3)^2+4 => y= 2x^2+12x+22
Y= 4(x+3)^2+2 => y= 4x^2+24x+38
Step-by-step explanation:
Where is the information, such as.. well um.. where is the line plot..
What was the distance in between?
Answer:
- sin(θ) = (-2√13)/13
- cos(θ) = (3√13)/13
- tan(θ) = -2/3
- sec(θ) = (√13)/3
- csc(θ) = (-√13)/2
- cot(θ) = -3/2
Step-by-step explanation:
The distance of the point from the origin is ...
r = √(3² +(-2)²) = √13
We know that ...
x = r·cos(θ) ⇒ cos(θ) = x/r = 3/√13 = (3√13)/13
y = r·sin(θ) ⇒ sin(θ) = y/r = -2/√13 = (-2√13)/13
tan(θ) = sin(θ)/cos(θ) = -2/3
sec(θ) = 1/cos(θ) = (√13)/3
csc(θ) = 1/sin(θ) = (-√13)/2
cot(θ) = 1/tan(θ) = -3/2
