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Naddika [18.5K]

4 years ago

12

How many different sets of polar coordinates can be given for a point, within one rotation? I thought it was infinite, but the g

iven options are 1, 2, 3, and 4.

2 answers:

Yanka [14]4 years ago

7 0

Answer:

the answer is 4

Step-by-step explanation:

so 1 rotation is like a circle 1 unit circle requires 4 quadrant to be in this is the most simplified i can get

liraira [26]4 years ago

3 0

Answer:

Solution : 4

Step-by-step explanation:

The question asks us how many polar coordinates are possible for one rotation. For one rotation there will be 4 polar coordinates, one present in each quadrant such that,

( r, theta ), ( r, theta ), ( - r, theta ), ( - r, theta )

Respectively if theta was q say,

( r, q ), ( r, - q ), ( - r, q ), ( - r, -q )

Therefore there are 4 sets of polar coordinates for one rotation, in each of the 4 quadrants.

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Solve for x. Round your answer to the nearest tenth if necessary.

diamong [38]

Answer:

x ≈ 17.5

Step-by-step explanation:

using Pythagoras' identity in the right triangle.

the square on the hypotenuse is equal to the sum of the squares on the other 2 sides , that is

x² = 9² + 15² = 81 + 225 = 306 ( take square root of both sides )

x = $\sqrt{306}$ ≈ 17.5 ( to the nearest tenth )

6 0

2 years ago

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Two particles move in the xy-plane. At time t, the position of particle A is given by x(t)=5t−5 and y(t)=2t−k, and the position

Ipatiy [6.2K]

Answer:

Part A) Not collide

Part B) k = 4

Part C) Particle B is moving fast.

Step-by-step explanation:

Two particles move in the xy-plane. At time, t

<u><em>Position of particle A:-</em></u>

$x(t)=5t-5$

$y(t)=2t-k$

<em><u>Position of particles B:-</u></em>

$x(t)=4t$

$y(t)=t^2-2t+1$

Part A) For k = -6

Position particle A, (5t-5,2t+6)

Position of particle B, $(4t,t^2-2t-1)$

If both collides then x and y coordinate must be same

Therefore,

For x-coordinate:

5t - 5 = 4t

t = 5

For y-coordinate:

$2t+6=t^2-2t-1$

$t^2-4t-7=0$

$t=-1.3,5.3$

The value of t is not same. So, k = -6 A and B will not collide.

Part B) If both collides then x and y coordinate must be same

For x-coordinate:

5t - 5 = 4t

t = 5

For y-coordinate:

$2t-k=t^2-2t-1$

Put t = 5

$10-k=25-10-1$

$k=4$

Hence, if k = 4 then A and B collide.

Part C)

Speed of particle A, $\dfrac{dA}{dt}$

$\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}$

$\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4$

Speed of particle B, $\dfrac{dB}{dt}$

$\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}$

$\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}$

At t = 5

$\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2$

Hence, Particle B moves faster than particle A

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4 years ago

Given the function f(x) = -2c + cx - x^2? and f^-1(5) = -1, find c.

ale4655 [162]

Answer:

c = - 2

Step-by-step explanation:

Given inverse function

$f^{-1}$ (5) = - 1 , then

f(- 1) = 5 , that is

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- 2c - c - 1 = 5

- 3c - 1 = 5 ( add 1 to both sides )

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3 years ago

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8_murik_8 [283]

Answer:

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Step-by-step explanation:

0.06 is one-tenth of the number that is 10 times bigger than 0.06.

0.06 * 10 = 0.6

0.06 is one-tenth of 0.6.

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3 years ago

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Pepsi [2]

Answer:

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3 years ago

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