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Answer:

#(x-11)/((x+5)(x-1)(x+1))#

Explanation:

First you want to expand both of the quadratic denominators. To do this for #x^2+4x-5# which is in standard form, you would need two factors of #-5# that add together to get #+4#. That would be the factors #5# and #-1#. So it expands to #(x+5)(x-1)#.

Now we have #4/((x+5)(x-1)) - 3/(x^2-1)#

You might recognize #x^2-1# as a perfect square that expands out to #(x+1)(x-1)#.

Now we have #4/((x+5)(x-1)) - 3/((x+1)(x-1))#

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 #4/((x+5)(x-1))# by the factor it's missing in its denominator which would be #(x+1)#.

#4/((x+5)(x-1))*(x+1)/(x+1) = (4(x+1))/((x+5)(x-1)(x+1)) = (4x+4)/((x+5)(x-1)(x+1))#

Next, we should multiply #3/((x+1)(x-1))# by the factor it's missing in its denominator which would be #(x+5)#.

#3/((x+1)(x-1)) * (x+5)/(x+5) = (3(x+5))/((x+5)(x-1)(x+1)) = (3x+15)/((x+5)(x-1)(x+1)#

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.

#((4x+4)-(3x+15))/((x+5)(x-1)(x+1)) = (4x+4-3x-15)/((x+5)(x-1)(x+1)) = (x-11)/((x+5)(x-1)(x+1))#

Answer:

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

Explanation:

Given the equation
#color(white)("XXX")5x+7y=77#

The x-intercept is the value of #x# when #y=0#
#color(white)("XXX")5x+7(0)=77#
#color(white)("XXX")rArr x= 15.4#

The y-intercept is the value of #y# when #x=0#
#color(white)("XXX")5(0)+7y=77#
#color(white)("XXX")rArr y= 11#

So the intercept points are at #(15.4,0)# and #(0,11)#

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]}

Answer:

It could be #a_n = (n^3+5n)/6#

Explanation:

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# using the first element of each sequence as a coefficient:

#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#

Answer:

#"vertex" -> (x,y) -> (2,1)#

Explanation:

#color(brown)("Introduction to idea of method.")#

Tony B

When the equation is in the form #a(x-b)^2+c# then #x_("vertex")=(-1)xx(-b)#

If the equation form had been #a(x+b)^2+c# then #x_("vertex")=(-1)xx(+b)#
#color(brown)(underline(color(white)(" ."))#

#color(blue)("To find "x_("vertex"))#

So for #y=3(x-2)^2+1 :#

#color(blue)(x_("vertex")=(-1)xx(-2)=+2)#
#color(brown)(underline(color(white)(" ."))#

#color(blue)("To find "y_("vertex"))#
Substitute +2 into the original equation to find #y_("vertex")#

So #y_("vertex")=3((2)-2)^2+1#

#color(blue)(y_("vertex")=0^2+1 = 1)#

#color(brown)("Also notice this value is the same as the constant of +1 that is in the"# #color(brown)("vertex form equation.")#
#color(brown)(underline(color(white)(" ."))#

Thus: #color(green)("vertex" -> (x,y) -> (2,1))#

#color(purple)("~~~~~~~~~~~~~~~~~~~ Foot note ~~~~~~~~~~~~~~")#

Suppose the equation had been presented in the form of:

#y=3x^2-12x+13#

write as #y= 3(x^2-4x) +13#

If we carry out the mathematical process of
#(-1/2) xx(-4)=+2 =x_("vertex")#

The -4 comes from the #-4x" in "(x^2-4x)#

#color(purple)(" ~~~~~~~~~~~~~~~~~~End Foot note~~~~~~~~~~~~~~~~~~~~~~~~")#

Answer:

#y^8/x^2#

Explanation:

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

#(x^-2y^-4x^3)^-2#

We can see that there are two #x# terms inside of the parenthesis. Lets combine those first. If we multiply two terms with exponents, the exponents add, in other words;

#x^a xx x^b = x^(a+b)#

Applying this to our case, we get;

#(x^((3-2))y^-4)^-2#

#(x^1y^-4)^-2#

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

#(x^a)^b = x^((a)(b))#

In our case, raising both the #x# term and the #y# term to #-2# we get;

#x^((-2)(1))y^((-2)(-4))#

#x^-2y^8#

Now we have one negative exponent and one positive exponent. We need to convert the #x# term to a positive exponent. To do that we will invert the term.

#x^-a = 1/x^a#

So to get rid of the #-# we will move the #x# term to the denominator.

#y^8/x^2#

Answer:

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)#

Explanation:

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 #3# to find:

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

Next note that the sum of the coefficients is zero. That is #2-3+1 = 0#. So #x=1# is a zero and #(x-1)# is a factor:

#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)#

Answer:

The vertex is #(-3/5,9)#.

Explanation:

#y=-25x^2-30x# is a quadratic equation in standard form, #ax^2+bx+c#, where #a=-25, b=-30, and c=0#. The graph of a quadratic equation is a parabola.

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 #a<0# opens downward.

Finding the Vertex
First determine the axis of symmetry, which will give you the #x# value. The formula for the axis of symmetry is #x=(-b)/(2a)#. Then substitute the value for #x# into the original equation and solve for #y#.

#x=-(-30)/((2)(-25))#

Simplify.

#x=(30)/(-50)#

Simplify.

#x=-3/5#

Solve for y.
Substitute the value for #x# into the original equation and solve for #y#.

#y=-25x^2-30x#

#y=-25(-3/5)^2-30(-3/5)#

Simplify.

#y=-25(9/25)+90/5#

Simplify.

#y=-cancel25(9/cancel25)+90/5#

#y=-9+90/5#

Simplify #90/5# to #18#.

#y=-9+18#

#y=9#

The vertex is #(-3/5,9)#.

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

Answer:

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

Explanation:

In ordinary Boolean logic, propositions are true or false.

In fuzzy logic you could consider propositions to have truth values in the range #[0, 1]# where #0# represents false, #1# represents 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(P)# for the truth value of a proposition #P#, then we might use the following rules:

#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 #P# and #Q# are independent? One option is to introduce ranges rather than single truth values. Another option is to introduce modal logical operators like necessity (#square#) and possibility (#diamond#). Even without 'fuzziness', modal logics can take quite a variety of forms.

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.

Answer:

.
#color(green)(x=0)# ..A different approach. Really this is the same thing as Tom's. The difference is that he has jumped steps and simplified using substitution.

Explanation:

Given: #color(brown)( (3x+2)/(3x-2) = (4x-7)/(4x+7)#

#color(blue)("'Getting rid' of the denominators")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)("Multiply both sides by "color(blue)(3x-2))#
#color(brown)( (3x+2)/(3x-2)color(blue)(xx(3x-2))= (4x-7)/(4x+7)color(blue)(xx(3x-2))#

#color(brown)( (3x+2)xx color(blue)((3x-2))/((3x-2))=((4x-7)color(blue)((3x-2)))/(4x+7))#

but #(3x-2)/(3x-2)# is another way of writing 1 giving:

#(3x+2) xx 1= ((4x-7)(3x-2))/(4x+7)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)("Multiply both sides by "color(blue)(4x+7))#

#color(brown)((3x+2) color(blue)(xx(4x+7))= ((4x-7)(3x-2))/(4x+7)color(blue)(xx(4x+7))#

#(3x+2)(4x+7)=(4x-7)(3x-2)xx((4x+7))/((4x+7))#

But# (4x+7)/(4x+7)# is another way of writing 1 giving:

#color(brown)((3x+2)color(blue)((4x+7))=(4x-7)color(black)((3x-2)))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)("Multiply out the brackets")#
#color(brown)(3xcolor(blue)((4x+7))+2color(blue)((4x+7)) =4x color(black)((3x-2))-7color(black)((3x-2))#

#12x^2+21x+8x+14=12x^2-8x-21x+14 #

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)("Collecting like terms")#

#(12x^2-12x^2)+(21x+8x+8x+21x)=14-14 #

#0x^2 +58x=0#

#color(green)(x=0)#

Answer:

DOMAIN: #x in (-oo,9)uu(9,+oo)#

RANGE: #y in (-oo,0)uu(0,+oo)#

Explanation:

#y=f(x)=k/g(x)#

Existence Condition is:

#g(x)!=0#

#:.x-9!=0#

#:.x!=9#

Then:

#F.E.#= Field of Existence=Domain: #x in (-oo,9)uu(9,+oo)#

#x=9# could be a vertical asymptote

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

  • #x rarr +-oo#

#lim_(x rarr -oo) f(x)=lim_(x rarr -oo) 5/(x-9)=5/-oo=0^-#

#lim_(x rarr +oo) f(x)=lim_(x rarr +oo) 5/(x-9)=5/(+oo)=0^+#

Then

#y=0# is a horizontal asymptote.

Indeed,

#f(x)!=0 AAx in F.E.#

  • #x rarr 9^(+-)#

#lim_(x rarr 9^-) f(x)=lim_(x rarr 9^-) 5/(x-9)=5/0^(-)=-oo#

#lim_(x rarr 9^+) f(x)=lim_(x rarr 9^+) 5/(x-9)=5/0^(+)=+oo#

Then

#x=9# it's a vertical asympote

#:. # Range of #f(x)#: #y in (-oo,0)uu(0,+oo)#