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statuscvo [17]
1 year 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]1 year 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]1 year 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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Answer:

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Los parámetros dados son;

El monto total gastado = 100 pesos

Sea el precio de las naranjas = x

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La cantidad pagada por las naranjas = 2 · y = x

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La cantidad total gastada = La cantidad pagada por las naranjas + La cantidad pagada por las bananas + La cantidad pagada por las manzanas

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z = y - 8 tenemos;

z = 27 - 8 = 19

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Step-by-step explanation:

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Or, Q =  50 g × 2260.0 J/g × 85°c

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Or, Q =  9,605 × 10³ joule

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