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Answer:

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

Explanation:

The first equation, #x - y + 4z = 6#, makes the following row in the augmented matrix:

#[ (1,-1,4,|,6) ]#

The second equation, #2x + z = 1#, makes the following row in the augmented matrix:

#[ (1,-1,4,|,6), (2,0,1,|,1) ]#

The third equation, #x + 5y + z = -9#, makes the following row in the augmented matrix:

#[ (1,-1,4,|,6), (2,0,1,|,1), (1,5,1,|,-9) ]#

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

#R_2-2R_1toR_2#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (1,5,1,|,-9) ]#

#R_3-R_1toR_3#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (0,6,-3,|,-15) ]#

#R_3-3R_2toR_3#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (0,0,18,|,18) ]#

#R_3/18#

#[ (1,-1,4,|,6), (0,2,-7,|,-11), (0,0,1,|,1) ]#

#R_2+7R_3toR_2#

#[ (1,-1,4,|,6), (0,2,0,|,-4), (0,0,1,|,1) ]#

#R_1-4R_2toR_1#

#[ (1,-1,0,|,2), (0,2,0,|,-4), (0,0,1,|,1) ]#

#R_2/2#

#[ (1,-1,0,|,2), (0,1,0,|,-2), (0,0,1,|,1) ]#

#R_1+R_2toR_1#

#[ (1,0,0,|,0), (0,1,0,|,-2), (0,0,1,|,1) ]#

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

#x = 0, y = -2, and z = 1#

Check:

#x - y + 4z = 6#
#2x + z = 1#
#x + 5y + z = -9#

#0 - -2 + 4(1) = 6#
#2(0) + 1 = 1#
#0 + 5(-2) + 1 = -9#

#6 = 6#
#1 = 1#
#-9 = -9#

This checks.

Answer:

# e/(e-1)^2=0.920674#

Explanation:

#sum_(n=1)^oo nx^n=x sum_(n=0)^oonx^(n-1)=x d/(dx)(sum_(n=0)^oo x^n)#

Here #x = e^-1 < 1# so

#sum_(n=0)^oo x^n=1/(1-x)# then

#sum_(n=1)^oo nx^n=x/(1-x)^2# for #abs x < 1# or

#sum_(n=1)^oo n(e^-1)^n=e^-1/(1-e^-1)^2 = e/(e-1)^2=0.920674#

Answer:

Please see below.

Explanation:

A conic equation of the type of #Ax^2+Bxy+Cy^2+Dx+Ey+F=0# is rotated by an angle #theta#, to form a new Cartesian plane with coordinates #(x',y')#, if #theta# is appropriately chosen, we can have a new equation without term #xy# i.e. of standard form.
http://philschatz.com/precalculus-book/contents/m49441.html
The relation between coordinates #(x,y)# and #(x'.y')# can be expressed as
#x=x'costheta-y'sintheta# and #y=x'sintheta+y'costheta#

or #x'=xcostheta+ysintheta# and #y=-xsintheta+ycostheta#

for this we need to have #theta# given by #cot2theta=(A-C)/B#

In the given case as equation is #x^2-3xy+4y^2+7=0#, we have #A=1#, #C=4# and #B=-3# and hence #cot2theta=(-3)/(-3)=1# i.e. #2theta=pi/4# and #theta=pi/8#.

Note that #sin(pi/8)=1/2sqrt(2-sqrt2)# and #cos(pi/8)=1/2sqrt(2+sqrt2)#.

Hence relation is give by #x=x'cos(pi/8)-y'sin(pi/8)# and #y=x'sin(pi/8)+y'cos(pi/8)# i.e.

#x=(x')/2sqrt(2+sqrt2)-(y')/2sqrt(2-sqrt2)# and #y=(x')/2sqrt(2-sqrt2)+(y')/2sqrt(2+sqrt2)#

Hence equation becomes

#((x')/2sqrt(2+sqrt2)-(y')/2sqrt(2-sqrt2))^2-3((x')/2sqrt(2+sqrt2)-(y')/2sqrt(2-sqrt2))((x')/2sqrt(2-sqrt2)+(y')/2sqrt(2+sqrt2))+4((x')/2sqrt(2-sqrt2)+(y')/2sqrt(2+sqrt2))^2+7=0#

Let us change #x'->x# and #y'->y# and simplifying we get

#x^2/4(2+sqrt2)+y^2/4(2-sqrt2)-(xy)/2sqrt2-3(x^2/4sqrt2-y^2/4sqrt2+(xy)/4(2+sqrt2)-(xy)/4(2-sqrt2))+4(x^2/4(2-sqrt2)+y^2/4(2+sqrt2)+(xy)/2sqrt2)+7=0#

or #x^2/4(2+sqrt2-3sqrt2+8-4sqrt2)+y^2/4(2-sqrt2+3sqrt2+8+4sqrt2)+(xy)/4(-2sqrt2-6-3sqrt2+6-3sqrt2+8sqrt2)+7=0#

or #x^2/4(10-6sqrt2)+y^2/4(10+6sqrt2)+7=0#

or #(5-3sqrt2)x^2+(5+3sqrt2)y^2+14=0#

You can follow a similar process for second equation #x^2+3xy-x+y=0#. Here #A=1#, #C=0# and #B=3# and hence #cot2theta=1/3# i.e. #2theta=cot^(-1)(1/3)#.

Answer:

Examine sequences of differences to find the formula...

Explanation:

Since the sum is of polynomial terms of degree #4#, the formula for the sum will be a polynomial of degree #5#.

If the terms of a sequence are given by a polynomial of degree #n#, then forming the sequence of differences, then of differences of differences #n# times will result in a constant sequence.

So the sequence of sums, being a polynomial of degree #5# will reduce to a constant sequence when we take differences #5# times.

To prepare, let us write down the first few powers of #4# (six terms will be just enough):

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

Now write the sequence of the first #6# sums:

#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 #n#th term of the sequence of sums:

#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)#

Answer:

#-12#

Explanation:

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 #y#.

#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 #x = 0#.

#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!

Answer:

Please see below.

Explanation:

Induction method is used to prove a statement. Most commonly, it is used to prove a statement, involving, say #n# where #n# represents the set of all natural numbers.

Induction method involves two steps, One, that the statement is true for #n=1# and say #n=2#. Two, we assume that it is true for #n=k# and prove that if it is true for #n=k#, then it is also true for #n=k+1#.

First Step #-# Now for #1^2+3^2+5^2+...+(2n-1)^2=n/3(4n^2-1)#, we know for #n=1#, we have #1^2=1/3(4*1^2-1)=1/3xx3=1# and for #n=2#, we have #1^2+3^2=2/3(4*2^2-1)=2/3xx15=10#.

Hence, given statement is true for #n=1# and #n=2#.

Second Step #-# Assume it is true for #n=k#, hence

#1^2+3^2+5^2+...+(2k-1)^2=k/3(4k^2-1)#

Now let us test it for #n=k+1# i.e.

#1^2+3^2+5^2+...+(2k-1)^2+(2k+1)^2#

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

= #k/3(4k^2+8k+4-1)-k/3(8k+4)+4k^2+4k+1#

= #k/3(4k^2+8k+4-1)-(8k^2)/3-(4k)/3+4k^2+4k+1#

= #(k+1)/3(4k^2+8k+4-1)-1/3(4k^2+8k+4-1)-(8k^2)/3-(4k)/3+4k^2+4k+1#

= #(k+1)/3(4k^2+8k+4-1)-color(red)((4k^2)/3)-color(blue)((8k)/3)-1-color(red)((8k^2)/3)-color(blue)((4k)/3)+color(red)(4k^2)+color(blue)(4k)+1#

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

Hence we see that the statement is true for #n=k+1# if it is true for #n=k#.

Hence #1^2+3^2+5^2+...+(2n-1)^2=n/3(4n^2-1)# is true for all values of #ninNN#

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