7.663013698630137 years ago

First you want to expand both of the quadratic denominators. To do this for

Now we have

You might recognize

Now we have

In order to add or subtract fractions we need to have like denominators. So in this case we can do some manipulation to meet that criterion like so:

First we can multiply

Next, we should multiply

By multiplying each fraction by a factor over itself it is the same as multiplying by one. Finally, we have the same denominator for each fraction and we can complete the subtraction in the numerator while, of course, the denominator will stay the same.

7.589041095890411 years ago

Plot the points for the x and y intercepts then draw a line through those points.

Given the equation

The x-intercept is the value of

The y-intercept is the value of

So the intercept points are at

If you plot these two points and draw a straight line through them, your graph should look something like:

graph{((x-15.4)^2 +y^2-0.1)(x^2+(y-11)^2-0.1)(5x+7y-77)=0 [-6.41, 25.63, -3.47, 12.55]}

7.589041095890411 years ago

It could be

You can always find a polynomial that matches a finite sequence like this one, but there are infinitely many possibilities.

Write out the original sequence:

#color(blue)(1),3,7,14#

Write out the sequence of differences:

#color(blue)(2),4,7#

Write out the sequence of differences of those differences:

#color(blue)(2),3#

Write out the sequence of differences of those differences:

#color(blue)(1)#

Having reached a constant sequence (!), we can write out a formula for

#a_n = color(blue)(1)/(0!)+color(blue)(2)/(1!)(n-1)+color(blue)(2)/(2!)(n-1)(n-2)+color(blue)(1)/(3!)(n-1)(n-2)(n-3)#

#=color(red)(cancel(color(black)(1)))+2n-color(red)(cancel(color(black)(2)))+color(red)(cancel(color(black)(n^2)))-3n+color(red)(cancel(color(black)(2)))+1/6n^3-color(red)(cancel(color(black)(n^2)))+11/6n-color(red)(cancel(color(black)(1)))#

#=(n^3+5n)/6#

7.413698630136986 years ago

When the equation is in the form

If the equation form had been

So for

Substitute +2 into the original equation to find

So

Thus:

Suppose the equation had been presented in the form of:

write as

If we carry out the mathematical process of

The -4 comes from the

7.410958904109589 years ago

We can use exponent rules to simplify this expression. Taking a look at the original function;

We can see that there are two

Applying this to our case, we get;

Now lets take a look at the exponent outside the parenthesis. Whenever we raise an exponent term to an exponent, we multiply the exponents.

In our case, raising both the

Now we have one negative exponent and one positive exponent. We need to convert the

So to get rid of the

7.405479452054794 years ago

Yes:

#y = 6x^3-9x+3#

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

#= 6(x-1)(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this later.

First separate out the common scalar factor

#y = 6x^3-9x+3 = 3(2x^3-3x+1)#

Next note that the sum of the coefficients is zero. That is

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

We can factor the remaining quadratic expression by completing the square and using the difference of squares identity...

#(2x^2-2x-1)#

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

#=2(x^2-x+1/4-3/4)#

#=2((x-1/2)^2-(sqrt(3)/2)^2)#

#=2((x-1/2)-sqrt(3)/2)((x-1/2)+sqrt(3)/2)#

#=2(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)#

Putting it all together:

#y = 6x^3-9x+3#

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

#= 6(x-1)(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)#

7.4 years ago

The vertex is

The vertex of a parabola is its minimum or maximum point. In this case it will be the maximum point because a parabola in which

**Finding the Vertex**

First determine the axis of symmetry, which will give you the

Simplify.

Simplify.

**Solve for y.**

Substitute the value for

Simplify.

Simplify.

Simplify

The vertex is

graph{y=-25x^2-30x [-10.56, 9.44, 0.31, 10.31]}

7.397260273972603 years ago

Fuzzy logic is a generalisation of Boolean logic with truth values between **true** and **false**.

In ordinary Boolean logic, propositions are **true** or **false**.

In fuzzy logic you could consider propositions to have truth values in the range **false**, **true** and any number in between represents a truth value strictly between **false** and **true**.

There are several different systems of fuzzy logic used for different purposes in different areas of mathematics. These different systems have different rules for the truth values of logical operations.

If we write

#v(not P) = 1-v(P)#

#v(P ^^ Q) = v(P)*v(Q)#

#v(P vv Q) = v(not ((not P) ^^ (not Q)))#

#= 1-(1-v(P))(1-v(Q))#

These rules correspond to the way that probabilities of two independent events would combine.

What can we do if we don't know whether *necessity* (*possibility* (

Yet another alternative is to introduce the notion of *relevance* into logic, which typically has the side effect of splitting simple *and*'s and *or*'s into intrinsic and extrinsic conjunctions and disjunctions. Combined with fuzziness, this can have a similar effect to the modal operators.

7.394520547945206 years ago

.

Given:

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

but

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

But

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

7.394520547945206 years ago

DOMAIN:

RANGE:

Existence Condition is:

Then:

To find the range we have to study the behavior for:

#x rarr +-oo#

Then

Indeed,

#x rarr 9^(+-)#

Then