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Featured 3 months ago

There are several ways to do it, using matrices. I prefer the use of an augmented matrix

The first equation,

The second equation,

The third equation,

Now, perform elementary row operations until, you obtain an identity matrix on the left.

We have an identity matrix on the left, therefore, the solution set is on the right:

Check:

This checks.

Featured 1 month ago

Here

Featured 3 weeks ago

Please see below.

A conic equation of the type of

The relation between coordinates

or

for this we need to have

In the given case as equation is

Note that

Hence relation is give by

Hence equation becomes

Let us change

or

or

or

You can follow a similar process for second equation

Featured 3 weeks ago

Examine sequences of differences to find the formula...

Since the sum is of polynomial terms of degree

If the terms of a sequence are given by a polynomial of degree

So the sequence of sums, being a polynomial of degree

To prepare, let us write down the first few powers of

#1, 16, 81, 256, 625, 1296#

Now write the sequence of the first

#color(blue)(1), 17, 98, 354, 979, 2275#

Next write the sequence of differences between pairs of successive terms:

#color(blue)(16), 81, 256, 625, 1296#

Write down the sequence of differences of those differences:

#color(blue)(65), 175, 369, 671#

Write down the sequence of differences of those differences:

#color(blue)(110), 194, 302#

Write down the sequence of differences of those differences:

#color(blue)(84), 108#

Write down the sequence of differences of those differences:

#color(blue)(24)#

We can now take the initial terms of each of these sequences as coefficients for a formula for the

#s_n = color(blue)(1)/(0!)+color(blue)(16)/(1!)(n-1)+color(blue)(65)/(2!)(n-1)(n-2)+color(blue)(110)/(3!)(n-1)(n-2)(n-3)+color(blue)(84)/(4!)(n-1)(n-2)(n-3)(n-4)+color(blue)(24)/(5!)(n-1)(n-2)(n-3)(n-4)(n-5)#

#color(white)(s_n) = 1+16n-16+65/2n^2-195/2n+65+55/3n^3-110n^2+605/3n-110+7/2n^4-35n^3+245/2n^2-175n+84+1/5n^5-3n^4+17n^3-45n^2+274/5n-24#

#color(white)(s_n) = 1/5n^5 + 1/2n^4 + 1/3n^3 - 1/30n#

#color(white)(s_n) = 1/30(6n^5 +15n^4+10n^3-n)#

Featured 3 weeks ago

There is a useful property that states that

#f(f^-1(x)) = x#

Therefore,

#f(f^-1(-12)) = -12#

I'm going to prove to you that this is true.

**Find the inverse of #y = (x+ 1)/(x - 2)# and prove that it is indeed the inverse. Use this to find the value of #y(y^-1(0)#**

Switching x and y values, we get:

#x = (y + 1)/(y - 2)#

Our goal here is to solve for

#x(y - 2) = y + 1#

#xy - 2x = y +1#

#xy - y= 1 + 2x#

#y(x - 1) = 1 + 2x#

#y = (2x + 1)/(x - 1)#

Now let's prove that this is indeed the inverse.

#y(y^-1) = ((2x + 1)/(x - 1) + 1)/((2x + 1)/(x - 1) - 2)#

#y(y^-1) = ((2x + 1 + x - 1)/(x - 1))/(((2x + 1 - 2x + 2))/(x - 1))#

#y(y^-1) = (3x)/3#

#y(y^-1) = x#

Now evaluate the inverse at

#y^-1(0) = (2(0) + 1)/(0 - 1) = -1#

Plug this into the initial function:

#y(-1) = (-1 + 1)/(0 - 2) = 0#

This is what we expected.

Hopefully you understand now!

Featured 1 week ago

Please see below.

**Induction method** is used to prove a statement. Most commonly, it is used to prove a statement, involving, say

Induction method involves two steps, One, that the statement is true for

**First Step**

Hence, given statement is true for

**Second Step**

Now let us test it for

=

=

=

=

=

=

Hence we see that the statement is true for

Hence

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