Answer:
27
Step-by-step explanation:
Answer:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
Step-by-step explanation:
Hello!
We need to determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°.
We know that the polar coordinate system is a two-dimensional coordinate. The two dimensions are:
- The radial coordinate which is often denoted by r.
- The angular coordinate by θ.
So we need to find r and θ. So we know that:
(1)
x = rcos(θ) (2)
x = rsin(θ) (3)
From the statement we know that (x, y) = (3, -3).
Using the equation (1) we find that:

Using the equations (2) and (3) we find that:
3 = rcos(θ)
-3 = rsin(θ)
Solving the system of equations:
θ= -45
Then:
r = 3\sqrt{2}[/tex]
θ= -45 or 315
Notice that there are two feasible angles, they both have a tangent of -1. The X will take the positive value, and Y the negative one.
So, the solution is:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
Answer: whats the question
Step-by-step explanation:
Step-by-step explanation:
The problem states that you have a linear function so expect your equation to have this form:
y = mx + b
where m is the slope and b is the y-intercept. You are also given two points: P1(5, 6) and P2(14, 60). Use these points to solve for the slope m.
m = (y2 - y1) / (x2 - x1) = (60 - 6)/(14 - 5)
= 54/9 = 6
So our equation now becomes
y = 6m + b
To solve for b, plug in the values of P1:
6 = 6(5) + b ---> b = -24
Therefore, our equation is
y = 6m - 24
The rest of the points are
(8, 24)
(11, 42)
Y/x=k
when y=9, then x= -3
9/-3=-3
⇒ k= -3