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3 years ago

Use pascal's triangle to expand the following binomial expression 1.(2k-1/3)⁶

Mathematics

1 answer:

Gwar [14]3 years ago

7 0

9514 1404 393

Answer:

64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

Step-by-step explanation:

The row of Pascal's triangle we need for a 6th power expansion is ...

1, 6, 15, 20, 15, 6, 1

These are the coefficients of the products (a^(n-k))(b^k) in the expansion of (a+b)^n as k ranges from 0 to n.

Your expansion is ...

1(2k)^6(-1/3)^0 +6(2k)^5(-1/3)^1 +15(2k)^4(-1/3)^2 +20(2k)^3(-1/3)^3 +...

15(2k)^2(-1/3)^4 +6(2k)^1(-1/3)^5 +1(2k)^0(-1/3)^6

= 64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

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3 years ago

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WITCHER [35]

Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but $x \ne 0$ </h3>

To write this in interval notation, we can say $(-\infty, 0) \cup (0, \infty)$ which is the result of poking a hole at 0 on the real number line.

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<h3>Range: Set of all real numbers</h3>

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Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation $x = \pm 4$

These are the locations where the blue hyperbolic curve crosses the x axis.

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Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.

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Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

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Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer: Oblique asymptote is y = 3x</h3>

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3 years ago

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Step-by-step explanation:

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3 years ago

Use pascal's triangle to expand the following binomial expression 1.(2k-1/3)⁶​

Use pascal's triangle to expand the following binomial expression 1.(2k-1/3)⁶