Show that the following rectangular form x^2+y^2=2xy can be expressed as 1 = sin(2) in polar form.

Mathematics

1 answer:

lara [203]2 years ago

7 0

Answer:

Step-by-step explanation:

substitute x = r*cos(θ), y = r*sin(θ) ==> r²(cos²(θ) + sin²(θ)) = 2r²cos(θ)sin(θ). Cancel the r² on both sides. On the left, use pythagorean identity cos²(θ) + sin²(θ) = 1. On the right apply double angle identity sin(2θ) = 2cos(θ)sin(θ).

This yields 1=sin(2θ). (I assume you meant to type sin(2θ) on the right hand side of the equation).

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if a pound of grapes cost $0.14 in 1960 and the price of grapes increased at the same rate as the CPI from 1960 to 1980 by appro

masya89 [10]

The CPI increased from 29.6 to 82.4.

The rate of increase, is:

82.4/29.6 = 2.78

Now multiply this rate to the $0.14 price of grapes from 1960.

$0.14 x 2.78 = $0.39

Now to get the amount of increase subtract the new price from the original price:

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3 years ago

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$