Featured 3 weeks ago

This seems a curious question to me, since normally you would use

Applying the Binomial theorem to

#(x+1)^8 = ""^8C_0 x^8 + ""^8C_1 x^7 + ""^8C_2 x^6 + ""^8C_3 x^5 + ""^8C_4 x^4 + ""^8C_5 x^3 + ""^8C_6 x^2 + ""^8C_7 x + ""^8C_8#

Taking the derivative

#8 * 7 * 6 (x+1)^5#

#= f^((3)) (x)#

#= 8 * 7 * 6 * ""^8C_0 x^5 + 7 * 6 * 5 * ""^8C_1 x^4 + 6 * 5 * 4 * ""^8C_2 x^3 + 5 * 4 * 3 * ""^8C_3 x^2 + 4 * 3 * 2 * ""^8C_4 x + 3 * 2 * 1 * ""^8C_5#

So:

#8 * 7 * 6#

#= 8 * 7 * 6 * ((color(blue)(0))+1)^5#

#= f^((3)) (0)#

#= 8 * 7 * 6 * ""^8C_0 (color(blue)(0))^5 + 7 * 6 * 5 * ""^8C_1 (color(blue)(0))^4 + 6 * 5 * 4 * ""^8C_2 (color(blue)(0))^3 + 5 * 4 * 3 * ""^8C_3 (color(blue)(0))^2 + 4 * 3 * 2 * ""^8C_4 (color(blue)(0)) + 3 * 2 * 1 * ""^8C_5#

#= 3 * 2 * 1 * ""^8C_5#

So:

#""^8C_5 = (8 * 7 * 6) / (3 * 2 * 1) = 56#

Alternatively, you might recognise that the method of constructing Pascal's triangle relates precisely to the values of the coefficients of

The row of Pascal's triangle starting

Featured 3 weeks ago

Recall that, for the **Hyperbola**

asymptotes are given by,

In our **Case,** the asymptotes are,

Next,

Multiplying by

By

With **Hyperbola** is,

Featured 2 weeks ago

**Solution set**

Alternate notation:

**Solve:**

**1) Examine the related equation to find the boundaries.**

or

The solutions to the related equation define boundary points that divide the number line into 5 parts, 3 regions and the 2 boundary points.

- Since the inequality is inclusive ("non-strict") the boundary points are in the solution set.

A graph of the number line, with solid dots on (-1), and (5) will help in thinking about the problem.

To check for the solution, **use test points, one from each region, in the inequality to see if they qualify:**

Try (-2), (0), and (6). It is best not to pick numbers that make the arithmetic difficult.

**In region,**

**In region,**

**In region,**

We must be careful in stating the solution inequalities. x can not be in both regions at once, since they are disconnected. There is a full region between the two regions in the solution, so we can't use "and", **we must use "or"**.

**Solution set**

Alternate notation:

**2-D Graph of the inequality:**

Featured 2 weeks ago

Perform the Gauss Jordan elimination on the augmented matrix

I have written the equations not in the sequence as in the question in order to get

Perform the folowing operations on the rows of the matrix

Thus

Featured 2 weeks ago

Please see below.

We know that

and

and hence

= **(A)**

and similarly

= **(B)**

Multipllying **(A)** and **(B)**, we get

Now what is coefficient of

While on LHS we have

On the RHS we get

Hence coefficient of

and hence

Featured 2 weeks ago

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "x^2+2=0rArrx^2=-2#

#"this has no real solutions hence there are no vertical"#

#"asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by the highest"#

#"power of x that is "x^2#

#f(x)=((2x)/x^2-3/x^2)/(x^2/x^2+2/x^2)=(2/x-3/x^2)/(1+2/x^2)#

#"as "xto+-oo,f(x)to(0-0)/(1+0)#

#rArry=0" is the asymptote"#

graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}

Featured 2 weeks ago

First find

The easiest way to find the inverse of

This is just:

Next switch the elements on the leading diagonal of

So you should have:

Divide each element by the determinant

Now:

Using

**Note we are post multiplying on both sides. This is important as Matrix multiplication is non-commutative.**

i.e.

**( In general )**

Featured 1 week ago

See below.

You can build a polynomial with as many inflexion points as needed using the truncated series expansion for

For instance, an example for

The next plot shows

And finally the plot for

NOTE

Depending on

Featured 1 week ago

Lets start by writing

in the form

Since

So the calculation essentially reduces to the determination of **not** diagonalizable. However, the characteristic equation for

and thus the Cayley-Hamilton theorem says that

so that the infinite series for

It is easy to calculate

Thus

Our solution is twice this matrix!

Featured yesterday

Let

What is

Subtracting latter from former, we get

or

=

=

=

=

and