How to solve for p=irt and solve for r
1 answer:
You might be interested in
Ypu would subtract Y from the other points Y
Then you would subtract X from the ither points X.
Then you would put the Y value over the X
Then simiplify
Then you would subtract X from the ither points X.
Then you would put the Y value over the X
Then simiplify
The intervals of the function on the graph are:
- Domain = (-∝, ∝)
- Range = (-∝, 2]
- Increases (-∝, 3]
- Decreases [3, ∝)
The function is given as:
f(x) = -2(x-3)² + 2
See attachment for the graph.
From the graph, we have;
Domain = (-∝, ∝)
Because it takes all real values as its input
Range = (-∝, 2]
Because it has a vertex of (3, 2) and the vertex is a maximum
Also, it increases on the interval (-∝, 3] and decreases on the interval [3, ∝)
Read more about function interval at:
#SPJ1
Answer: cos(Θ) = (√15) / 4
Explanation:
The question states:
1) sin(Θ) = 1/4
2) 0 < Θ < π / 2
3) find cos(Θ)
This is how you solve it.
1) Use the fundamental identity (in this part I use α instead of Θ, just for facility of wirting the symbols, but they mean the same for the case).
2) From which you can find:
3) Replace sin(α) with 1/4
=>
=>
4) Given that the angle is in the first quadrant, you know that cosine is positive and the final answer is:
cos(Θ) =
.
And that is the answer.
Explanation:
The question states:
1) sin(Θ) = 1/4
2) 0 < Θ < π / 2
3) find cos(Θ)
This is how you solve it.
1) Use the fundamental identity (in this part I use α instead of Θ, just for facility of wirting the symbols, but they mean the same for the case).
2) From which you can find:
3) Replace sin(α) with 1/4
=>
=>
4) Given that the angle is in the first quadrant, you know that cosine is positive and the final answer is:
cos(Θ) =
And that is the answer.