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Yuliya22 [10]
3 years ago
6

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

+ (x−1)2+ (x−1)3+ (x−1)4+ (x−1)5. NOTE: This exercise gives a method for writing a polynomial in powers of x−1 instead of powers of x. This may be useful, for example, if you are interested in the values of the polynomial near x=1 .
Mathematics
1 answer:
DENIUS [597]3 years ago
5 0

Compute the necessary values/derivatives of f(x) at x=1:

f(1)=3

f'(1)=2

f''(1)=-12

f'''(1)=-12

f^{(4)}(1)=48

f^{(5)}(1)=120

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) f(x) at x=1 by

T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5

T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5

###

Another way of doing this would be to solve for the coefficients a,b,c,d,e,g in

f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5

by expanding the right hand side and matching up terms with the same power of x.

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A sandbox has a perimeter of 21 1/2 feet. If the length is 5 1/4 feet, what is the area of the sand box?
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The width will be the difference between this and the length, hence 5 1/2 ft.

The area is the product of length and width.
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4 0
3 years ago
A parabola and its directrix are shown on the graph. On a coordinate plane, a parabola opens upward. It has a vertex at (0, 0) a
sukhopar [10]

Answer:

The focus of the parabola is at the point (0, 2)

Step-by-step explanation:

Recall that the focus of a parabola resides at the same distance from the parabola's vertex, as the distance from the parabola's vertex to the directrix, and on the side of the curve's concavity. In fact this is a nice geometrical property of the parabola and the way it can be constructed base of its definition: "All those points on the lane whose distance to the focus equal the distance to the directrix."

Then, the focus must be at a distance of two units from the vertex, (0,0), on in line with the parabola's axis of symmetry (x=0), and on the positive side of the y-axis (notice the directrix is on the negative side of the y-axis. So that puts the focus of this parabola at the point (0, 2)

5 0
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(17 points) <br> Determine if line AB is tangent to the circle
zlopas [31]

1. If AB is tangent to the circle, then

18^2=(2\cdot7.2)^2+10.8^2

We have

18^2=324

(2\cdot7.2)^2+10.8^2=324

so AB is indeed tangent to the circle.

2. The unlabeled leg is another radius of the circle so it has length 8. Then if AB is tangent to the circle,

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so that cannot be a right triangle and AB is not tangent to the circle.

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Viefleur [7K]

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the number is 8

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8-3=5

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expression: 6(x-3)>24

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I am pretty sure x is 2 and y is 7
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