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Answer:

Please see below.

Explanation:

Given that the axis is #x = -3#, we know the vertex is at #x = -3#, so the equation can be written in the form

#y = a(x+3)^2 +k#

Now use the two points given to get:

From #(-1,3)#, we have #3 = a(2)^2+k# so #4a+k = 3# and #k = 3-4a#

From #(2,-5)#, we get #-5 = a(5)^2+k# so #25a+k=-5#

Replace #k# in the last equation with #3-4a# to get

#25a+(3-4a) = -5#, so

#21a = -8# and # a = -8/21#

Using #k = 3-4a# again we get #k = 3-(-32/21) = 63/21+32/21 = 95/21#.

The equation is #y = -8/21(x+3)^2 + 95/21#

If course, there are other ways to write the answer.

graph{(y+8/21(x+3)^2-95/21)(x+3)((x+1)^2+(y-3)^2-0.02)((x-2)^2+(y+5)^2-0.02)=0 [-12, 12, -6, 6]}

Answer:

The solution is #((x),(y),(z))=((1),(2),(6))#

Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

#((5,5,-6,:,-21),(-2,1,-3,:,-18),(-1,6,-6,:,-25))#

#R1larr-R3# and #R3larrR1#

#((1,-6,6,:,25),(-2,1,-3,:,-18),(5,5,-6,:,-21))#

#R3larr2R3+5R2#

#((1,-6,6,:,25),(-2,1,-3,:,-18),(0,15,-27,:,-132))#

#R2larrR2+2R1#

#((1,-6,6,:,25),(0,-11,9,:,32),(0,15,-27,:,-132))#

#R3larr11R3-15R2#

#((1,-6,6,:,25),(0,-11,9,:,32),(0,0,-162,:,-972))#

#R3larr(R3)/(-162)#

#((1,-6,6,:,25),(0,-11,9,:,32),(0,0,1,:,6))#

#R2larrR2-9R3#

#((1,-6,6,:,25),(0,-11,0,:,-22),(0,0,1,:,6))#

#R2larr(R2)/(-11)#

#((1,-6,6,:,25),(0,1,0,:,2),(0,0,1,:,6))#

#R1larrR1+6R2-6R3#

#((1,0,0,:,1),(0,1,0,:,2),(0,0,1,:,6))#

Answer:

Steve M

Explanation:

Let # omega=-3+3i #, and let #z^8 = omega#

First, we will put the complex number into polar form:

# |omega| = sqrt((-3)^2+3^2) =3sqrt(2)#
# theta =arctan(3/(-3)) = arctan(-1) = -pi/4#
# => arg \ omega = pi/2+pi/4 = (3pi)/4 #

So then in polar form we have:

# omega = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4)) #

Wolfram Alpha

We now want to solve the equation #z^8=omega# for #z# (to gain #8# solutions):

# z^8 = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4)) #

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of #2pi#, so we can equivalently write (incorporating the periodicity):

# z^8 = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi)) \ \ \ n in ZZ #

By De Moivre's Theorem we can write this as:

# z = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8) #
# \ \ = (3sqrt(2))^(1/8)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8) #
# \ \ = 3^(1/8)2^(1/16) (cos(((3pi)/4+2npi)/8) + isin(((3pi)/4+2npi)/8))#
# \ \ = 3^(1/8)2^(1/16) (cos theta + isin theta ) #

Where:

# theta =((3pi)/4+2npi)/8 = ((3+8n)pi)/32#

And we will get #8# unique solutions by choosing appropriate values of #n#. Working to 3dp, and using excel to assist, we get:
Steve M

After which the pattern continues (due the above mentioned periodicity).

We can plot these solutions on the Argand Diagram:

Wolfram Alpha

In linear algebra, one studies the notion of linear vector spaces over different scalar fields.

So assuming that you are familiar with the notion of a linear vector space, let us consider an LVS #V(F)#.

Then for two elements #u# and #v# in #V#, we can associate a scalar (in the field #F#) such that the following properties hold,

1) Complex conjugate of #(u,v) = (v,u) #

2) #(u,u) >= 0# and the equality holds only for #u = 0#

3) For four vectors #u, v, w, z# in #V# and two arbitrary scalars #a# and #b# in #F#, #(a(u + v), b(w + z)) = bar a b [(u,w) + (u,z) + (v,w) + (v,z)]#

Where #bar a# denotes complex conjugation.

If these properties hold, we call #(u,v)# an inner product of #u# and #v# and #V# is then an inner product space.

However, there is no general recipe for defining the inner product of two vectors (or functions). We just have to make sure that the properties 1-3 are satisfied.

For an example, an inner product of two wave functions #psi# and #phi# in Quantum mechanics is defined as,

#< psi|phi> = int bar psi phi d tau# where integration is over all volume of the space and #bar psi# denotes complex conjugate of #psi#.

One can easily check if the properties 1-3 hold.

Answer:

#(x^4-2x^3-4x^2+2x+3) -: (x^2+2x+1) = x^2-4x+3#, with no remainder.

Explanation:

Set up the long division like this:

#color(magenta)(x^2)+2x+1bar(|"  "color(magenta)(x^4)-2x^3-4x^2+2x+3)#

Divide #color(magenta)(x^4/x^2)#, giving #color(red)(x^2)#; put this quotient above the #x^4#:

#color(white)(x^2+2x+1bar(|"  "color(red)(x^2)#
#x^2+2x+1bar(|"  "x^4-2x^3-4x^2+2x+3)#

Multiply #color(red)(x^2) xx (x^2+2x+1)#, giving #color(blue)(x^4+2x^3+x^2)#; put this product below the #x^4-2x^3-4x^2#:

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2)#
#x^2+2x+1bar(|"  "x^4-2x^3-4x^2+2x+3)#
#color(white)(x^2+2x+1bar(|"  "color(blue)(x^4+2x^3+color(white)(1)x^2))#

Subtract #(x^4-2x^3-4x^2)-(color(blue)(x^4+2x^3+x^2))#, giving #color(orange)(–4x^3-5x^2)#; draw a line under #color(blue)(x^4+2x^3+x^2)# and write this difference below the line:

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2)#
#x^2+2x+1bar(|"  "x^4-2x^3-4x^2+2x+3)#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4"  ")color(orange)(-4x^3-5x^2)"          ")#

Copy the #color(green)(2x)# from the dividend down below this line:

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2)#
#x^2+2x+1bar(|"  "x^4-2x^3-4x^2+color(green)(2x)+3)#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4)-4x^3-5x^2" "color(green)(+2x))#

Repeat this process twice, dividing the latest leading term below your line by the leading #x^2# from the divisor:

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2-color(red)(4x))#
#color(magenta)(x^2)+2x+1bar(|"  "x^4-2x^3-4x^2+2x+3)#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4)-color(magenta)(4x^3)-5x^2+2x)#

...

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2-color(red)(4x))#
#color(red)(x^2+2x+1)bar(|"  "x^4-2x^3-4x^2+2x+3)#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4)-4x^3-5x^2+2x)#
#color(white)(x^2+2x+1|bar("     "color(blue)(-4x^3-8x^2-4x)))#

...

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x)#
#x^2+2x+1bar(|"  "x^4-2x^3-4x^2+2x+color(green)(3))#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4)-4x^3-5x^2+2x)#
#color(white)(x^2+2x+1|bar("     "color(black)(-4x^3-8x^2-4x)))#
#color(white)(x^2+2x+1|"        ")bar(color(white)(-4x^3+)color(orange)(3x^2+6x)+color(green)(3))#

...

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+color(red)3)#
#color(magenta)(x^2)+2x+1bar(|"  "x^4-2x^3-4x^2+2x+3)#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4)-4x^3-5x^2+2x)#
#color(white)(x^2+2x+1|bar("     "color(black)(-4x^3-8x^2-4x)))#
#color(white)(x^2+2x+1|"        ")bar(color(white)(-4x^3+)color(magenta)(3x^2)+6x+3)#

...

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+color(red)(3))#
#color(red)(x^2+2x+1)bar(|"  "x^4-2x^3-4x^2+2x+3)#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4)-4x^3-5x^2+2x)#
#color(white)(x^2+2x+1|bar("     "color(black)(-4x^3-8x^2-4x)))#
#color(white)(x^2+2x+1|"        ")bar(color(white)(-4x^3+)3x^2+6x+3)#
#color(white)(x^2+2x+1|"        "bar(color(blue)("            "3x^2+6x+3)#

...

#color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+3)#
#x^2+2x+1bar(|"  "x^4-2x^3-4x^2+2x+3)#
#color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))#
#color(white)(x^2+2x+1|)bar("  "color(white)(x^4)-4x^3-5x^2+2x)#
#color(white)(x^2+2x+1|bar("     "color(black)(-4x^3-8x^2-4x)))#
#color(white)(x^2+2x+1|"        ")bar(color(white)(-4x^3+)3x^2+6x+3)#
#color(white)(x^2+2x+1|"        "bar(color(black)("            "3x^2+6x+3)#
#color(white)(x^2+2x+1|"                       ")bar(color(white)(3x^2+6x+color(orange)0)#

Answer:

#2047#

Explanation:

Note that #1024 = 2^10#

This is a geometric series, with common ratio #2#, initial term #1# and #11# terms.

The general term of a geometric series can be written:

#a_n = ar^(n-1)#

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

Then we find:

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

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

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

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

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

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

In our example #N=11#, #a=1#, #r=2# and

#sum_(n=1)^N a_n = (color(blue)(1)(1 - color(blue)(2)^color(blue)(11)))/(1-color(blue)(2)) = 2^11-1 = 2047#

Alternatively, in this particular example there is a simpler solution...

Note that:

#1+(1+2+4+8+...+1024) = 2+2+4+8+...+1024#

#color(white)(1+(1+2+4+8+...+1024)) = 4+4+8+16+...+1024#

#color(white)(1+(1+2+4+8+...+1024)) = 8+8+16+32+...+1024#

#color(white)(1+(1+2+4+8+...+1024) =) vdots#

#color(white)(1+(1+2+4+8+...+1024)) = 1024+1024#

#color(white)(1+(1+2+4+8+...+1024)) = 2048#

So:

#1+2+4+8+...+1024 = 2048-1 = 2047#

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