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d1i1m1o1n [39]
3 years ago
10

Evan is going to invest $2,800 and leave it in an account for 12 years. Assuming the

Mathematics
2 answers:
anyanavicka [17]3 years ago
8 0

Answer:

5000=2000/1 hshshsjsnndjskbdjs

Dominik [7]3 years ago
8 0

Answer:

r = 4.95046%              

r ≈ 4.95%                      

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Find all polar coordinates of point P where P = ordered pair 3 comma negative pi divided by 3 .
jek_recluse [69]

Answer:

The all polar coordinates of P are:

(3 , -π/3) , (3 , 5π/3) , (-3 , 2π/3) , (-3 , -4π/3)

Step-by-step explanation:

* Lets study the polar coordinates of a point

- In polar coordinates there is an infinite number of coordinates

 for a given point.

- The polar coordinates of a point (x , y) is (r , θ), where

  r = √ ( x2 + y2 )

  θ = tan-1 ( y / x )

# Ex: the following four points are all coordinates for the same point.

* (5 , π/3) = (5 , −5π/3) = (−5 , 4π/3) =(−5 , −2π/3)

- These four points only represent the coordinates of the point without  

  rotating more than once

- So the point (r,θ) can be represented by any of the following

  coordinate pairs  (r , θ + 2π n) and (−r , θ + (2n + 1) π), where n is

  any integer.

* Now lets solve the problem

∵ P = (3 , -π/3)

∵ (r , θ + 2πn)

∴ r = 3 an d Ф = -π/3

- let n = 1

∴ P = (3 , -π/3 + 2π)

∴ P = (3 , 5π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = 0

∴ P = (-3 , -π/3 + (2×0 + 1) π)

∴ P = (-3 , -π/3 + (0 + 1) π)

∴ P = (-3 , -π/3 + π)

∴ P = (-3 , 2π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = -1

∴ P = (-3 , -π/3 + (2(-1) + 1) π)

∴ P = (-3 , -π/3 + (-2 + 1) π)

∴ P = (-3 , -π/3 + -π) = (-3 , -4π/3)

∴ P = (-3 , -4π/3)

5 0
3 years ago
Please help me with 1,2,3,4,5 and show me the steps and the answers do not send me link please
lina2011 [118]

Answer:

I did the first one on paper

Step-by-step explanation:

i also sent one page of my personal notes on right triangle Pythagoerean thereom I hope this helps

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2 years ago
Help with 1 and 2 please
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love your nails, can u bring it a bit closer tho?

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WORLD PEACE EVERY WHERE
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yes Amen and we also need Jesus

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