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Answer:

The answer is #=-2/(x-1)+5/(2x+1)+3/(x-2)#

Explanation:

We must factorise the denominator

Let #f(x)=2x^3-5x^2+x+2#

#f(1)=2-5+1+2=0#

Therefore, #(x-1)# is a factor of #f(x)#

To find the other factors, we have to do a long division

#color(white)(aaaa)##2x^3-5x^2+x+2##color(white)(aaaa)##∥##x-1#

#color(white)(aaaa)##2x^3-2x^2##color(white)(aaaaaaaaaaa)##∥##2x^2-3x-2#

#color(white)(aaaaaa)##0-3x^2+x#

#color(white)(aaaaaaaa)##-3x^2+3x#

#color(white)(aaaaaaaaaaa)##0-2x+2#

#color(white)(aaaaaaaaaaaaa)##-2x+2#

#color(white)(aaaaaaaaaaaaaa)##-0+0#

Therefore,

#2x^3-5x^2+x+2=(x-1)(2x^2-3x-2)#

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

Now, we can perform our decomposition into partial fractions

#(7x^2-12x+11)/(2x^3-5x^2+x+2)=(7x^2-12x+11)/((x-1)(2x+1)(x-2))#

#=A/(x-1)+B/(2x+1)+C/(x-2)#

#=(A(2x+1)(x-2)+B(x-1)(x-2)+C(2x+1)(x-1))/((x-1)(2x+1)(x-2))#

Therefore,

#7x^2-12x+11=A(2x+1)(x-2)+B(x-1)(x-2)+C(2x+1)(x-1)#

Let #x=1#, #=>#, #6=-3A#, #=>#, #A=-2#

Let #x=2#, #=>#, #15=5C#, #=>#, #C=3#

Coefficients of #x^2#, #=>#, #7=2A+B+2C#

#B=7-2A-2C=7+4-6=5#

So,

#(7x^2-12x+11)/(2x^3-5x^2+x+2)=-2/(x-1)+5/(2x+1)+3/(x-2)#

Answer:

#12#

Explanation:

The general term of a geometric series can be represented by the formula:

#a_n = a r^(n-1)#

where #a# is the initial term and #r# the common ratio.

We find:

#(1-r) sum_(n=1)^N ar^(n-1) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a + color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - ar^N#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a(1 - r^N)#

Dividing both ends by #(1-r)# we find:

#sum_(n=1)^N ar^(n-1) = (a(1 - r^N))/(1-r)#

If #abs(r) < 1# then #lim_(N->oo) r^N = 0# and we find:

#sum_(n=1)^oo ar^(n-1) = lim_(N->oo) (a(1 - r^N))/(1-r) = a/(1-r)#

In the given example, #a=6# and #r=1/2#. So #abs(r) < 1# and:

#sum_(n=1)^oo 6*(1/2)^(n-1) = 6/(1-1/2) = 12#

Answer:

#{ (a=1), (b=9), (c=27), (d=23) :}#

Explanation:

Given:

#{ (-5 = -64a+16b-4c+d), (-4 = -27a+9b-3c+d), (-3 = -8a+4b-2c+d), (4 = -a+b-c+d) :}#

Consider the function:

#f(x) = ax^3+bx^2+cx+d#

Note that the given system of equations is equivalent to:

#{ (f(-4) = -5), (f(-3) = -4), (f(-2) = -3), (f(-1) = 4) :}#

Since the sampling points are at equal intervals of size #1#, we can conveniently find a formula for a cubic function satisfying this system by examining differences...

Start with the original values:

#color(blue)(-5), -4, -3, 4#

Write down the sequence of differences between successive terms:

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

Write down the sequence of differences between successive terms:

#color(blue)(0), 6#

Write down the sequence of differences between successive terms:

#color(blue)(6)#

Then we can write down a formula for #f(x)# using the initial term of each of these sequences (see footnote):

#f(x) = color(blue)(-5)/(0!)+color(blue)(1)/(1!)(x+4)+color(blue)(0)/(2!)(x+4)(x+3)+color(blue)(6)/(3!)(x+4)(x+3)(x+2)#

#color(white)(f(x)) = -5+x+4+x^3+9x^2+26x+24#

#color(white)(f(x)) = x^3+9x^2+27x+23#

So:

#{ (a=1), (b=9), (c=27), (d=23) :}#

#color(white)()#
Footnote

Note that:

#color(purple)(1/(0!))_(color(white)(1/1))# is a constant function taking the value #1# when #x = -4#

#color(purple)(1/(1!)(x+4))_(color(white)(1/1))# is a linear function taking the value #0# when #x = -4# and #1# when #x = -3#

#color(purple)(1/(2!)(x+4)(x+3))_(color(white)(1/1))# is a quadratic function taking the value #0# when #x = -4# or #x = -3# and the value #1# when #x = -2#

#color(purple)(1/(3!)(x+4)(x+3)(x+2))_(color(white)(1/1))# is a cubic function taking the value #0# when #x = -4#, #x = -3# or #x = -2# and the value #1# when #x = -1#

So as we add suitable multiples of these functions in turn, we get a sequence of polynomials of increasing degree that match each of the sample points in turn. The suitable multiples are the differences we found.

Answer:

#+-1/sqrt6(-2,-1,1).#

Explanation:

Let us recall that, for vectors #veca and vecb#, their Vector

(Cross) Product , i.e., #veca xx vecb# is Orthogonal to both

#vec a and vec b.#

The Desired Unit Vectors, then, can be obtained by,

#+-(vecaxxvecb)/||(vecaxxvecb)||#.

Now, with #veca=i-j+k=(1,-1,1), &, vecb=4j+4k=4(0,1,1)#, we have,

#veca xx vecb=|(i,j,k),(1,-1,1),(0,4,4)|=4|(i,j,k),(1,-1,1),(0,1,1)|#

#=4(-2,-1,1)#,

#rArr ||(vecaxxvecb)||=4sqrt{(-2)^2+(-1)^1+1^2}=4sqrt6.#

Hence, the desired vectors are, #(+-4(-2,-1,1))/(4sqrt6), or, #

#+-1/sqrt6(-2,-1,1).#

Enjoy Maths.!

Answer:

Minor: #|(5,4),(4,-1)|color(white)("XXX")#Co-factor: #-|(5,4),(4,-1)|#

Explanation:

The minor of a determinant is created by deleting all elements in the same row or column as the selected element:
#|(5,4,color(red)(cancel(color(black)2))), (color(red)(cancel(color(black)(1))),color(red)(cancel(color(black)(7))),color(red)(cancel(3))), (4,-1,color(red)(cancel(color(black)(1))))| =|(5,4),(4,-1)|#

The co-factor of a determinant is created by multiplying the minor by #(-1)^(r+c)# where #r# is the row number of the selected element and #c# is the column number of the selected element.
In this case #color(red)(3)# is in row #2# column #3#.
(The top left is always row #1#, column #1#)
So for this example
#color(white)("XXX")(-1)^(r+c)=(-1)^(2+3)=(-1)^5=-1#
and the co-factor is
#color(white)(""XXX")-1 * |(5,4),(4,-1)|#

Answer:

See below

Explanation:

#A = ((1,0),(9,4))#

To diagonalise, we need the matrix's eigenvectors. And .... the matrix needs to offer up 2 distinct eigenvectors in order to be diagonalisable.

Handy trick: For a triangular matrix, the eigenvalues are its diagonal entries.

But we can show that here by going the long way round: they are also the solutions to the characteristic equation, which for a #2 times 2# is:

#lambda^2 - Tr(A) lambda + det A = 0#

#implies lambda^2 - (1+4) lambda + (1*4 - 9*0) = 0#

#implies lambda = 1, 4#, the diaginal entries!

The eigenvectors (#mathbf alpha_(1,2)#) follow from the basic geometric eigenvalue idea that #A mathbf alpha_(1,2) = lambda_(1,2) \ mathbf alpha_(1,2)#:

  • # lambda = 1#

#x_1 + 0 x_2 = x_1#
#9 x_1 + 4 x_2 = x_2#

From the first equation, #x_1# can be anything: we choose #x_1 = 1#. This means from the 2nd equation that #x_2 = -3#. Thus:

#mathbf alpha_1 = ((1),(-3))#

  • # lambda = 4#

#x_1 + 0 x_2 = 4x_1#
#9 x_1 + 4 x_2 = 4 x_2#

From the first equation, #x_1 = 0# is the only solution. This means from the 2nd equation that #x_2 = 1#. Thus:

#mathbf alpha_2 = ((0),(1))#

The diagonalisation matrix P is then simply:

#P = (mathbf alpha_1, mathbf alpha_2) = ((1,0),(-3,1))#

The matrix D is a diagonal matrix that has the eigenvalues as it only non-zero entries:

#D = ((1,0),(0,4))#

We are asserting that:

#A = PDP^(-1)#

Now, we have everything apart from #P^(-1)#, which is:

#P^(-1) = ((1,0),(3,1))#

So our assertion here is that:

# ((1,0),(9,4)) = ((1,0),(-3,1)) ((1,0),(0,4)) ((1,0),(3,1)) #

Try it and see.

You can then do some seriously cool stuff with matrix. There is a point to diagonalisation :)

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