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Ipatiy [6.2K]
3 years ago
14

What is the y-intercept of the function, represented by the table of values below? x: -2,1,2,4,7 y: 14,8,6,2,-4

Mathematics
1 answer:
anastassius [24]3 years ago
7 0

Answer:

The y-intercept is at y = 10.

Step-by-step explanation:

It will be between y = 14 and y = 7 because the corresponding x values are -2 and 1.

An increase of 3 units of x  gives a decrease of 6 units of y fro the above values.

Then an increase of 1 ( 1 to 2) in x gives decrease in y of 2 (8 to 6). The other values show the same pattern.

So very unit increase in x,  the y values change by -2.

So  from x = -2 to 0 is +2 units for x and this will be -4 units for y  so the y-intercept ( when x = 0) will be at y = 14-4 = 10

y-intercept  is (0,10).


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Find the solution of 4x2y′′−4x2y′+y=0,x>04x2y″−4x2y′+y=0,x>0 of the form y1=xr(1+c1x+c2x2+c3x3+⋯)
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\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

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GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

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From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

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