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°)
the amount you walk by the hour?
x axis- distance
y axis- hours
:))
The product of a scalar and a matrix is found by multiplying each element of the matrix by the scalar. Multiply each element by -4.
... [4 -12 -24 12]
Answer:
The length of river frontage for each lot are 96.55 ft. 98.85 ft, 101.15 ft and 103.45 ft.
Step-by-step explanation:
See the attached diagram.
The river frontage of 400 ft will be divided into 84 : 86 : 88 : 90 for each lot as AP, BQ, CR, DS and ET all are parallel.
Therefore, PQ : QR : RS : ST = 84 : 86 : 88 : 90 = 42 : 43 : 44 : 45
Let, PQ = 42x, QR = 43x, RS = 44x and ST = 45x
So, (42x + 43x + 44x + 45x) = 400
⇒ 175x = 400
⇒ x = 2.2988.
So, PQ = 42x = 96.55 ft.
QR = 43x = 98.85 ft.
RS = 44x = 101.15 ft and
ST = 45x = 103.45 ft
(Answer)
Answer:
the gcf of 16, 20, and 14 is 4