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2

Answer:

#""^8C_5 = 56#

Explanation:

This seems a curious question to me, since normally you would use #""^8C_5 = (8!)/(3!5!)# to find a binomial coefficient, but let's give it a try.

Applying the Binomial theorem to #f(x) = (x+1)^8# we have:

#(x+1)^8 = ""^8C_0 x^8 + ""^8C_1 x^7 + ""^8C_2 x^6 + ""^8C_3 x^5 + ""^8C_4 x^4 + ""^8C_5 x^3 + ""^8C_6 x^2 + ""^8C_7 x + ""^8C_8#

Taking the derivative #3# times, we find:

#8 * 7 * 6 (x+1)^5#

#= f^((3)) (x)#

#= 8 * 7 * 6 * ""^8C_0 x^5 + 7 * 6 * 5 * ""^8C_1 x^4 + 6 * 5 * 4 * ""^8C_2 x^3 + 5 * 4 * 3 * ""^8C_3 x^2 + 4 * 3 * 2 * ""^8C_4 x + 3 * 2 * 1 * ""^8C_5#

So:

#8 * 7 * 6#

#= 8 * 7 * 6 * ((color(blue)(0))+1)^5#

#= f^((3)) (0)#

#= 8 * 7 * 6 * ""^8C_0 (color(blue)(0))^5 + 7 * 6 * 5 * ""^8C_1 (color(blue)(0))^4 + 6 * 5 * 4 * ""^8C_2 (color(blue)(0))^3 + 5 * 4 * 3 * ""^8C_3 (color(blue)(0))^2 + 4 * 3 * 2 * ""^8C_4 (color(blue)(0)) + 3 * 2 * 1 * ""^8C_5#

#= 3 * 2 * 1 * ""^8C_5#

So:

#""^8C_5 = (8 * 7 * 6) / (3 * 2 * 1) = 56#

Alternatively, you might recognise that the method of constructing Pascal's triangle relates precisely to the values of the coefficients of #(a+b)^n# and hence deduce that the appropriate entry in Pascal's triangle must be the binomial coefficient.

The row of Pascal's triangle starting #1, 8,...# contains the coefficients for #(a+b)^8#, with the coefficient of #a^3b^5# being #56#, a.k.a. #""^8C_5# ...

enter image source here

1

Answer:

#x^2/9-y^2/16=1#.

Explanation:

Recall that, for the Hyperbola # S : x^2/a^2-y^2/b^2=1#, the

asymptotes are given by, #y=+-b/a*x;" where, "a,b gt 0#.

In our Case, the asymptotes are, #y=+-4/3*x#.

#:. b/a=4/3...........................................................................(1)#.

Next, #(3sqrt2,4) in S rArr (3sqrt2)^2/a^2-4^2/b^2=1#.

#:. 18/a^2-16/b^2=1#.

Multiplying by #b^2#, we get, #18*b^2/a^2-16=b^2#.

By #(1)#, then, #18*(4/3)^2-16=b^2, i.e., b^2=16#.

#:. b=4 (because, b >0), and, (1) rArr a=3, or, a^2=9#.

With #a^2=9, and b^2=16#, the reqd. eqn. of the Hyperbola is,

#x^2/9-y^2/16=1#.

1

Answer:

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

Explanation:

Solve:
# x^2 − 4x ≥ 5 #

1) Examine the related equation to find the boundaries.
# x^2 − 4x ≥ 5 #
# x^2 − 4x - 5 = 0 #
# ( x - 5 ) ( x + 1 ) = 0 #
# ( x - 5 ) = 0 #
# x = 5 #
or
# ( x + 1 ) = 0 #
# x = -1 #

The solutions to the related equation define boundary points that divide the number line into 5 parts, 3 regions and the 2 boundary points.

  • Since the inequality is inclusive ("non-strict") the boundary points are in the solution set.

A graph of the number line, with solid dots on (-1), and (5) will help in thinking about the problem.

To check for the solution, use test points, one from each region, in the inequality to see if they qualify:

Try (-2), (0), and (6). It is best not to pick numbers that make the arithmetic difficult.

In region, # -oo < x < -1 #
# ( -2 )^2 − 4( -2) ≥ 5 " ?"#
# 4 + 8 ≥ 5 " ?"#
# 12 ≥ 5 " YES"#

In region, # -1 < x < 5 #
# ( 0 )^2 − 4( 0 ) ≥ 5 " ?" #
# 0 ≥ 5 " NO" #

In region, # 5 < x < oo #
# ( 6 )^2 − 4( 6 ) ≥ 5 " ?" #
# 36 − 24 ≥ 5 " ?" #
# 12 ≥ 5 " YES" #

We must be careful in stating the solution inequalities. x can not be in both regions at once, since they are disconnected. There is a full region between the two regions in the solution, so we can't use "and", we must use "or".

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

2-D Graph of the inequality:
enter image source here

1

Answer:

#r=5#, #s=3# and #t=7#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,1,1,|,15),(1,0,1,|,12),(0,1,1,|,10))#

I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R2larrR2-R1#

#A=((1,1,1,|,15),(0,-1,-0,|,-3),(0,1,1,|,10))#

#R1larrR1+R2#; #R3larrR3+R2#

#A=((1,0,1,|,12),(0,-1,-0,|,-3),(0,0,1,|,7))#

#R1larrR1-R3#

#A=((1,0,0,|,5),(0,-1,-0,|,-3),(0,0,1,|,7))#

#R2larr(R2)*(-1)#

#A=((1,0,0,|,5),(0,1,0,|,3),(0,0,1,|,7))#

Thus #r=5#, #s=3# and #t=7#

1

Answer:

Please see below.

Explanation:

We know that

#(a+b)^n=C_0^na^n+C_1^na^(n-1)b+C_2^na^(n-2)b^2+ ⋯ + C_n^nb^n#

and #r^(th)# term is #C_r^na^rb^(n-r)# and #C_r^n=(n!)/(r!(n-r)!)#

and hence #(1+x)^n=C_0 1^n+C_1 1^(n-1)*x+C_2 1^(n-2)x^2+ ⋯ + C_nx^n#

= #(1+x)^n=C_0 1^n+C_1x+C_2x^2+ ⋯ + C_nx^n# (A)

and similarly #(x+1)^n=C_0 x^n+C_1 x^(n-1)*1+C_2 x^(n-2)1^2+ ⋯ + C_n1^n#

= #(x+1)^n=C_0x^n+C_1x^(n-1)+C_2x^(n-2)+ ⋯ +C_n1# (B)

Multipllying (A) and (B), we get #(1+x)^n(x+1)^n# on the LHS and multiplication of expansion on the RHS.

Now what is coefficient of #x^(n+r)# in this.

While on LHS we have #(1+x)^(2n)# and coefficient of #x^(n+r)# is #(2n!)/((n+r)!(2n-n-r)!)=(2n!)/((n+r)!(n-r)!)#

On the RHS we get #x^(n+r)#, when #C_0x^n# is multiplied by #C_rx^r#; #C_1x^(n-1)# is multiplied by #C_(r+1)x^(r+1)#; #C_2x^(n-2)# is multiplied by #C_(r+2)x^(r+2)# and so on till we get #C_nx^n# multipled by #C_(r+n)x^(n-(r+n)#.

Hence coefficient of #x^(n+r)# is

#C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)#

and hence #C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)=((2n)!)/((n+r)!(n-r)!)#

1

Answer:

#"horizontal asymptote at "y=0#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "x^2+2=0rArrx^2=-2#

#"this has no real solutions hence there are no vertical"#
#"asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by the highest"#
#"power of x that is "x^2#

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

#"as "xto+-oo,f(x)to(0-0)/(1+0)#

#rArry=0" is the asymptote"#
graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}

2

Answer:

#bb(Y)=[(-11/16,17/16),(-1/4,3/4)]#

Explanation:

First find #bb(A^-1)#

The easiest way to find the inverse of #bb(A)# is to find the determinant of #bb(A)#:

This is just:

#(3xx6)-(1xx2)=16#

Next switch the elements on the leading diagonal of #bb(A)# and change the signs of the elements on the non-leading diagonal of #bb(A)#

So you should have:

#[(6,-2),(-1,3)]#

Divide each element by the determinant #bb(16)#:

#[(6/16,-2/16),(-1/16,3/16)]=[(3/8,-1/8),(-1/16,3/16)]#

#bb(A^-1)=[(3/8,-1/8),(-1/16,3/16)]#

Now:

#bb(YA)+bb(B)=bb(C)#

#bb(YA)=bb(C-B)#

Using #bb(A^-1)#

#bb(YA A^-1)=bb((C-B)A^-1)#

#bb(Y)=bb((C-B)A^-1)#

Note we are post multiplying on both sides. This is important as Matrix multiplication is non-commutative.

i.e.

#bb(AB)!=bb(BA)# ( In general )

#bb(C-B)=[(3,4),(2,6)]-[(4,-1),(2,2)]=[(-1,5),(0,4)]#

#:.#

#bb(Y)=[(-1,5),(0,4)][(3/8,-1/8),(-1/16,3/16)]=[(-11/16,17/16),(-1/4,3/4)]#

#bb(Y)=[(-11/16,17/16),(-1/4,3/4)]#

1

Answer:

See below.

Explanation:

You can build a polynomial with as many inflexion points as needed using the truncated series expansion for #sinx = sum_(k=0)^n (-1)^kx^(2k-1)/((2k-1)!)# added to a line with convenient gradient or as

#p_n(x,m) = sum_(k=0)^n (-1)^kx^(2k-1)/((2k-1)!)+ m x#

For instance, an example for #m = -2# and #n = 21# with exactly #11# inflexion points.

enter image source here

The next plot shows #d/(dx)p_(21)(x,-2)#. As we can observe #d/(dx)p_(21)(x,-2) = 0# does not have real roots, then #p_(21)(x,-2)# has not relative maxima/minima.

enter image source here

And finally the plot for #d^2/(dx^2)p_(21)(x,-2)# showing the inflexion points location. as the roots of #d^2/(dx^2)p_(21)(x,-2) = 0# . Here we can count exactly #11# inflexion points,

enter image source here

NOTE

Depending on #n# the sign for #m# can be positive or negative, and #n > 4#. It is left as an exercise to determine the connection for the #m# sign with the #n# value as well as the dependency between the sought inflection points number, with the #n# value.

4

Answer:

#((2,2,3,13/3),(0,2,2,3),(0,0,2,2),(0,0,0,2))#

Explanation:

Lets start by writing
# ( ( ln(2), 1, 1, 1 ), ( 0, ln(2), 1, 1), ( 0, 0, ln(2), 1 ), ( 0, 0, 0, ln(2) ) )#

in the form #ln(2) I_4 +A#, where #I_4# is the #4\times 4# identity matrix and #A# is the upper triangular matrix

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

Since #I_4# commutes with all #4 times 4# matrices in general, and #A# in particular, the exponential matrix is simply

#exp(ln(2) I_4 +A) = exp(ln(2) I_4) exp(A) = e^{ln(2)}I_4 exp(A) = 2 exp(A)#

So the calculation essentially reduces to the determination of #e^A#. Note that the standard approach using diagonalization does not work here, since #A# is not diagonalizable. However, the characteristic equation for #A# is easily seen to be
#lambda^4 =0#
and thus the Cayley-Hamilton theorem says that
#A^4 = 0#
so that the infinite series for #e^A# truncates :

#e^A = I+A+1/{2!} A^2 + 1/{3!}A^3#

It is easy to calculate

#A^2 = ((0,0,1,2), (0,0,0,1), (0,0,0,0), (0,0,0,0)) and A^3 = ((0,0,0,1),(0,0,0,0),(0,0,0,0),(0,0,0,0))#

Thus
#e^A = ((1,1,3/2,13/6),(0,1,1,3/2),(0,0,1,1),(0,0,0,1)) #

Our solution is twice this matrix!

2

Answer:

#S_n=3-(2n+3)/2^n#

Explanation:

Let #S_n=1/2+3/4+5/8+.................."n terms"#

What is #n^(th)# term of the series? Well numerators are in arithmatic sequence and numerator of #n^(th)# term is #2n-1# and denominator is in geometric series with #n^(th)# term being #2^n#. Hence #n^(th)# term of given series is #(2n-1)/2^n# and series is

#S_n=1/2+3/4+5/8+7/16+..................+(2n-1)/2^n# and then

#1/2S_n=1/4+3/8+5/16+..................+(2n-1)/2^(n+1)#

Subtracting latter from former, we get

#S_n-1/2S_n=1/2+2/4+2/8+....+2/2^n-(2n-1)/2^(n+1)#

or #1/2S_n=1/2+2/4[1+1/2+....+1/2^(n-2)]-(2n-1)/2^(n+1)#

= #1/2+1/2(1-1/2^(n-1))/(1-1/2)-(2n-1)/2^(n+1)#

= #1/2+1-1/2^(n-1)-(2n-1)/2^(n+1)#

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

= #3/2-(2n+3)/2^(n+1)#

and #S_n=3-(2n+3)/2^n#