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statuscvo [17]
2 years ago
11

The following curve passes through (3,1). Use the local linearization of the curve to find the approximate value of y at x=2.8.

...?
Mathematics
2 answers:
Vesnalui [34]2 years ago
6 0

ANSWER:

y ≈ (-10/19)*(2.8 - 3) + 1 = 1 2/19 ≈ 1.105

STEP-BY-STEP EXPLANATION:

y = (2x+13)/(2x^2+1)

y' = ((2x^2+1)*(2) - (2x+13)(4x)) / (2x^2+1)^2

at x=3, this is

y'(3) = (19*2 - 19*12)/19^2 = -10/19

So, your linearization is

y ≈ (-10/19)*(x-3) + 1

At x=2.8, this is

y ≈ (-10/19)*(2.8 - 3) + 1 = 1 2/19 ≈ 1.105

Artist 52 [7]2 years ago
3 0

d/dx (2 x^2 y + y = 2x + 13) 4xy + 2x^2 y' + y' = 2 4xy + y'(2x^2 + 1) = 2 y' = (2- 4xy)/(2x^2 +1)

<span>ow we can use this in a linear equation for a slope Ty = -5x/8 +5(3)/8 +8/8 = -5x/8 +(15+8)/8 = -5x/8 +23/8 this will gives us an approximation at x=2.8 now</span>

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A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave
erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

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C. open at both ends.

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Step-by-step explanation:

Given:

Length of the pipe, L = 120 cm

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We have to find whether the pipe is open,closed or open-closed or none.

Note:

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For one side open and one side closed pipe:

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⇒ \lambda_2=\frac{480}{3}                                  ⇒  \lambda_3=\frac{480}{5}

⇒ \lambda_2=160\ cm                           ⇒   \lambda_3=96\ cm  

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