Question

Find the distance between P1(3, –195°) and P2(–4, –94°) on the polar plane. Round your answer to the nearest thousandth.

I got 5.75 as my answer. Is this correct?

Answer 1

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Answer 2

Yes it is correct

Answer 3

Revamped
P1(3, –195°)
x = r cos ∝ = 3 cos(-195) = 3(-0.966)
x = -2.9
y = r sin α = 3 sin(-195) = 3(0.259)
y = 0.8
P1(x, y) = (-2.9, 0.8)

P2(–4, –94°)
x = -4 cos(-94) = -4(-0.0698) = 0.28
y = -4 sin(-94) = -4(0.998) = -3.99
P2(0.28, -3.99)

(-2.9, 0.8) and (0.28, -3.99)
P1 to P2 distance is
d = √(y2 – y1)² + (x2 – x1)²
= √((-3.99 – 0.8)² + (0.28 + 2.9)²)
= √(-4.79² + 3.18²) = √33.0565
= 5.75

Answer 4

Thanks! Can you help me?
@PyroYolka

Answer 5

Awesome thank-you so much!

Answer 6

P1(3, –195°)
x = r cos ∝ = 3 cos(-195) = 3(-0.966)
x = -2.9
y = r sin α = 3 sin(-195) = 3(0.259)
y = 0.8
P1(x, y) = (-2.9, 0.8)

P2(–4, –94°)
x = -4 cos(-94) = -4(-0.0698) = 0.28
y = -4 sin(-94) = -4(0.998) = -3.99
P2(0.28, -3.99)

(-2.9, 0.8) and (0.28, -3.99)
P1 to P2 distance is
d = √(y2 – y1)² + (x2 – x1)²
= √((-3.99 – 0.8)² + (0.28 + 2.9)²)
= √(-4.79² + 3.18²) = √33.0565
= 5.75

Answer 7

hmmm....

Answer 8

Can you help me with one more?

Answer 9

@PyroYolka

Answer 10

srry my os crashed close this question and open a new one :)