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goldfiish [28.3K]
3 years ago
14

In a unit circle, ø = 2pi radians What is the terminal point?

Mathematics
2 answers:
allochka39001 [22]3 years ago
5 0

Answer:

(1,0)

Step-by-step explanation:

Luba_88 [7]3 years ago
4 0

Answer:

The terminal point is (1, 0) ⇒ D

Step-by-step explanation:

In the unit circle, Ф is the angle between the terminal side and the positive part of the x-xis

  • The terminal point on the positive part of the x-axis is (1, 0),which means Ф = 0 or 2π radians and cosФ = 1, sinФ = 0
  • The terminal point on the positive part of the y-axis is (0, 1),which means Ф = \frac{\pi }{2} radians and cosФ = 0, sinФ = 1
  • The terminal point on the negative part of the x-axis is (-1, 0),which means Ф = π radians and cosФ = -1, sinФ = 0
  • The terminal point on the negative part of the y-axis is (0, -1),which means Ф = \frac{3\pi }{2} radians and cosФ = 0, sinФ = -1

In a unit circle

∵ Ф = 2π radians

→ By using the 1st rule above

∴ The terminal point is (1, 0)

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1/2 of 4/3 = 1/2 of ______thirds =______ thirds
Leya [2.2K]

Answer:

four, two

Step-by-step explanation:

1/2 of 4/3 is half of four thirds. Half of four is two, so it would be two thirds

<em>I hope this helps! :) </em>

4 0
3 years ago
How do I solve this
poizon [28]

Answer:

8x^{-2} a^{7}

Step-by-step explanation:

combine by putting together

3 0
4 years ago
What is the area of a rectangle with vertices at (−3, −1) , (1, 3) , (3, 1) , and (−1, −3) ?
77julia77 [94]
The coordinates of the rectangle are (-3,-1), (1,3), (3,1) and (-1,-3).

Draw the given rectangle as shown in the figure below.

Calculate the length of the rectangle as
a = √(4² + 4²) = 4√2
Calculate the width of the rectangle as
b = √(2² + (-2)²) = 2√2
Calculate the area of the rectangle as
A = a*b = (4√2) * (2√2) = 16

Answer: 16

4 0
3 years ago
The area of an isosoles trapezoid is 45cm squared, and it's height is 5cm. Find the perimeter if the length of a leg is 2/3 of t
Schach [20]

Answer:

Perimeter = 42 cm.

Step-by-step explanation:

The  2 legs are equal in length as it is an isosceles trapezoid.

Area of  trapezoid  = (h/2)(a + b)  where a and b are the parallel bases.

So

(5/2) (a + b) = 45

a + b = 45 * 2/5 = 18 cm.

We are given that the length of a leg (L) is 2/3 * sum of the bases, so:

L = 2/3(a + b)

  = 2/3 * 18

   = 12 cm.

The perimeter = a + b + 2L

= 18 * 2*12

= 42 cm.

3 0
2 years ago
Suppose r(t) = (et cos t)i + (et sin t)j. Show that the angle between r and a never changes. What is the angle?
tatyana61 [14]

Answer:

Angle = Ф = cos^{-1}(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

Step-by-step explanation:

Given vector r(t) = e^{t}cost i + e^{t}sint j

As we know that,

velocity vector = v = \frac{dr}{dt}

Implies that

velocity vector = (e^{t} cost - e^{t} sint)i + (e^{t} sint - e^{t}cost )j

As acceleration is velocity over time so:

acceleration vector = a = \frac{dv}{dt}

Implies that

vector a =

(e^{t}cost - e^{t}sint - e^{t}sint - e^{t}cost )i + ( e^{t}sint + e^{t}cost + e^{t}cost - e^{t}sint )j

vector a = (-2e^{t}sint) i + ( 2e^{t}cost)j

Now scalar product of position vector r and acceleration vector a:

r. a = . \\

r.a = -2e^{2t}sintcost + 2e^{2t}sintcost

r.a = 0

Now, for angle between position vector r and acceleration vector a is given by:

cosФ = \frac{r.a}{|r|.|a|} = \frac{0}{|r|.|a|} = 0

Ф = cos^{-1}(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

4 0
3 years ago
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