Answer:
The required vector parametric equation is given as:
r(t) = <3cost, 3sint>
For 0 ≤ t ≤ 2π
Step-by-step explanation:
Given that
f(x, y) = <2y, -sin(y)>
Since C is a cirlce centered at the origin (0, 0), with radius r = 3, it takes the form
(x - 0)² + (y - 0)² = r²
Which is
x² + y² = 9
Because
cos²β + sin²β = 1
and we want to find a vector parametric equations r(t) for the circle C that starts at the point (3, 0), we can write
x = 3cosβ
y = 3sinβ
So that
x² + y² = 3²cos²β + 3²sin²β
= 9(cos²β + sin²β) = 9
That is
x² + y² = 9
The vector parametric equation r(t) is therefore given as
r(t) = <x(t), y(t)>
= <3cost, 3sint>
For 0 ≤ t ≤ 2π
Let's solve ~
[ according to given figure ]
Correct choice is D
Answer:
The slope of the line through (−9,6) and (−6,−9) is −5
Step-by-step explanation:
Hope it helps
Unusual notation. I won't fuss with it.
a. We have isosceles PRT, so angle RPT = angle RTP.
By the definition of angle bisector, angle MTP = angle MTF, and angle MPT = angle MPU.
We have m angle RTP = m angle MTF + m angle MTP = 2 m angle MTP
Similarly, m angle RPT = 2 m angle MPT
2 m angle MTP = 2 m angle MPT
angle MPT = angle MPT
That's the first part.
b. That makes MPT isosceles.
c. 2x+124=180
2x = 56
x = 28 degrees
MTP = 28 degrees
d. We have angle RPT=angle RTP=56 so PRT=180-2(56)=68 degrees
PUT = 180 - UTP - UPT = 180 - 28 - 56 = 96 degrees
Bad drawing, PUT looks acute.
angle PRT = 68 degrees, angle PUT = 96 degrees
