Questions on Algebra: Complex Numbers answered by real tutors!

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I need help solving: z=6-4i
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Solving for what?
Is i a variable? Or sqrt(-1) ?

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(3+2i)(3+2i)
===============
= 9 + 12i + 4i^2
If i is
then = 9 + 12i - 4
= 5 + 12i

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Combine like terms. Since 7x and -3x both contain an x, you can subtract them normally. 7x-3x=4x
You're then left with 4x=80
You want want to get x alone.
Divide each side by 4.
You're left with x=20.

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the general equation is ___ y = ax^2 + bx + c

substituting the given points will form a system of equations that can be solved for the a, b, and c coefficients

(-1,0) ___ (0) = a(-1)^2 + b(-1) + c

(0,-5) ___ (-5) = a(0)^2 + b(0) + c

(2,3) ___ (3) = a(2)^2 + b(2) + c

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Start with the given equation.


Notice that the quadratic is in the form of where , , and


Let's use the quadratic formula to solve for "x":


Start with the quadratic formula


Plug in , , and


Square to get .


Multiply to get


Subtract from to get


Multiply and to get .


Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


Break up the fraction.


Reduce.


or Break up the expression.


So the solutions are or
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6(7-w)+5(3+2w)


42-6w+15+10w


4w+57


So 6(7-w)+5(3+2w) simplifies to 4w+57
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Subtract 5 from both sides of equation

x = 7

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By a change of variables, determine the indefinite integral as follows:
integral (x^2 - 3x^4)^1/2 dx
=====================================
The integrand can be factored as:

Let
Then
So the integral can be written as

The antiderivative of
Substituting back the expression for y gives

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Firstly, we know that (a+b)^p=a^p+b^p in every field of characteristic p. Therefore s(a+b)=s(a)+s(b). Obviously, s(ab)=s(a)s(b). So s(a) is a homomorphism. The only way for s(a) to equal 0 is if a is 0, so the kernel of the homomorphism is {0} and therefore the map is one to one. Finally, we know that s (a) is onto since the field involved is finite. Thus, the Frobenius map is an automorphism.

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The number of complex solutions of a polynomial is always equal to the degree of the polynomial (this is called the fundamental theorem of algebra).

You can find the possible rational roots quite easily using the rational root theorem. The possible rational roots of a polynomial are in the form where p is a factor of the constant term and q is a factor of the leading coefficient. This theorem can easily be proven using modular arithmetic.

There isn't much of a way to find possible "real" roots. However, you do know that if P(a) is negative and P(b) is positive, there exists at least one real zero between a and b. This is due to the intermediate value theorem, which states that for a continuous function f(x) between a and b, every number between f(a) and f(b) has at least one x-value in the domain.

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You have to know 2 things:

1. How to change from a+bi form to trig form r(cosq + i sinq)

use tanq =  and r = 

2.  How to raise a complex number in trig form using DeMoivre's
   formula:

[r(cosq + i sinq)]n = rn[cos(nq) + i·sin(nq)]


(2+2squareroot3i)^6


1. To change from 2+2i form to trig form r(cosq + i·sinq)

use tanq =  = 
therefore q = 60°

and r =  =  =  =  = 4

So 2 + 2i = 4(cos60° + i·sin60°) 


2. To raise this complex number in trig form using DeMoivre's
   formula:

[4(cos60° + i·sin60°)]6 = 46[cos(6·60°) + i·sin(6·60)] =
4096(cos360° + i·sin360°) = 4096(cos0° + i·sin0°) = 
4096(1+i·0) = 4096(1+0) = 4096(1) = 4096.

answer: 4096.

Edwin

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Int(u dv)=uv-Int(v du) {this is integration by parts: I won't prove it, but it isn't difficult to show}. Let u=ln(x) and dv=x^(1/2)dx. Then du=dx/x and v=(2/3)x^(3/2). Thus Int(x^(1/2)ln(x)dx)=(ln(x))*(2/3)x^(3/2)-Int((2/3)x^(3/2)(1/x)dx)=(ln(x))*(2/3)x^(3/2)-Int((2/3)x^(1/2)dx)=(ln(x))*(2/3)x^(3/2)-(4/9)x^(3/2)=MESS

[**MESS=(2/9)x^(3/2)(3ln(x)-2)]

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First decompose the given fraction into its partial fractions:
Factor the denominator:
now we set:
Now add the fractions on the right side.
Now since the denominators are equal, the numerators must be equal, so we can set:
This must be true for all x so it is true for the x-values that make:

so and
so
So we can solve for A, B, and C by letting x = 0, then x = -2, and x = 3.
After some standard algebra, you'll find that:
, , and we can now write the partial fraction:

Now when you integrate these partial fractions you'll get the answers:

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The polynomial has a double root (-2).

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Adding complex numbers is quite simple to show graphically, since complex numbers add just like vectors (add the individual components). The sum of the complex numbers is simply the resultant vector formed by the other numbers.

Multiplying/dividing is a little more difficult to show graphically. Perhaps you could represent two complex numbers z1, z2 by



(note that Euler's formula states that e^(ix) = cos x + i sin x, so you can represent any complex number in the above form).

Then,



Graphically, this means that the radius is numerically equal to the product of the radii of r1 and r2, and the angle formed by z1z2 is equal to the sum of the angles formed by z1 and z2. Representing division graphically is similar; think of division as multiplying by 1 over the complex number. Or, find some way to "undo" the multiplication.

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Nickels -------x
dimes ----------y
5x+10y =215...................(1)
Nickels------x+28
Dimes---------2y
5(x+28)+10*2y= 465
5x+140+20y=465
5x+20y=325--------------------(2)

subtract (2) from (1)
-10y=-110
/-10
y=11 dimes
plug y in (1) to get value of x
x=21 nickels
CHECK
21*5+10*11=215

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A: Given then



By Euler's formula, (this is proven using Taylor series -- you'll see these in calculus) so we have



, done.

B: Rewrite using Euler's formula:

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LCD = (x+3) (x-1)
Multiply the equation by LCD

=







(x+6)(2x+1)=0
x=-6 OR -0.5
m.ananth@hotmail.ca

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Multiply top and bottom by




check answer:




OK

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So
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Start with the given expression.


FOIL.


Multiply.


Replace with . Note: .


Multiply.


Combine like terms.


So .


So the expression is now in standard form where and
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Start with the given equation.


Get every term to the left side.


Combine like terms.


Notice that the quadratic is in the form of where , , and


Let's use the quadratic formula to solve for "x":


Start with the quadratic formula


Plug in , , and


Square to get .


Multiply to get


Subtract from to get


Multiply and to get .


Take the square root of to get .


or Break up the expression.


or Break up the fraction for each case.


or Reduce.


or Simplify.


So the solutions are or
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(2-3i)-(5-6i)


2-3i-5+6i


(2-5)+(-3i+6i)


-3+3i
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x^4-2x^3-x^2-4x-6=0
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Try 1, 2, 3 & 6, the factors of 6, both plus and minus.

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F(x) = 4x^3 - 5x + 2x^(1/2). Determine the slope of the tangent when x = 4?
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F'(x) = 12x^2 - 5 + x^(-1/2)
F'(4) = 12*16 - 5 + 1/2
Slope @ x = 4 = 187.5

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This is a calculus 1 question.

You need to find the derivative of F(x).

I'm looking for F'(x). So, I will do this by differenting termwise.

(4x^3)' = 12x^2

(-5x)' = -5

(2x^1/2)' = x^(-1/2)

We can write x^(-1/2) as 1/sqrt{x}.

So, F'(x) = 12x^2 + 1/sqrt{x} - 5

To find the slope of the tangent line, we now replace every x you see with 4 and simplify.

F'(4) = 12(4)^2 + 1/sqrt{4} - 5

F'(4) = 12*16 + 1/2 - 5

F'(4) = 375/2

The slope of the tangent line is 375/2.

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The impedance Z (or opposition to the flow of electricity) of the metal is related to the voltage V and current I by the equation, where Z, V, and I are complex numbers.
Calculate Z for the given values of V and I.
V=10+20i I=3+7i
:
Z = V/I
:
Z =
:
multiply by the conjugate of the denominator over itself
Z = * = = = =

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x5-3x4-8x3-8x2-9x-5=0

The possible rational zeros are ± the factors of 5, 
so they are ±1, and ±5

We try the easiest one x=1

1|1 -3  -8  -8  -9  -5
 |   1  -2 -10 -18 -27 
  1 -2 -10 -18 -27 -32

No that gives remaider -32, not 0, so 1 is not a solution

We try the next easiest one x=-1

-1|1 -3  -8 -8 -9 -5
  |  -1   4  4  4  5 
   1 -4  -4 -4 -5  0

That gives remaider 0, so -1 is a solution and so we have
factored the polynomial on the left as:

(x + 1)(x4 - 4x³ - 4x² - 4x - 5) = 0 

So now we try to factor that polynomial in the second
parentheses:

It also has the same set of possible solutions, so we try -1
again (there is no use to try 1) 

-1|1 -4 -4 -4 -5 
  |  -1  5 -1  5 
   1 -5  1 -5  0

That gives remaider 0, so -1 is a DOUBLE solution 
and so we have further factored the polynomial on the left:

(x + 1)(x + 1)(x³ - 5x² + x - 5) = 0

We can now factor the polynomial in the third parentheses by
grouping:

               x³ - 5x² + x - 5
              x²(x - 5) + 1(x - 5)
              (x - 5)(x² + 1)

So we have now factored the original left side as

(x + 1)(x + 1)(x - 5)(x² + 1) = 0

Now we use the zero-factor principle and set each factor = 0

x + 1 =  0,  x + 1 = 0,  x - 5 = 0, x² + 1 = 0
    x = -1       x = -1      x = 5.     x² = -1
                                         x = 
                                         x = ±i

So the solutions are  -1, -1, 5, i, -i.

-1 was a solution twice, so we call it a "double root".

Edwin

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=

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1. Write the expression in simplified radical form:

multiply by the conjugate of the denominator over itself
* = = = =
we can cancel -7 into 28 and -7; results

:
:
3. Solve the equation. Show your check, and then write the solution set.
x-1 =
x - 1 - 10 =
square both sides
(x-11)^2 = 6x
x^2 - 22x + 121 = 6x
x^2 - 22x - 6x + 121 = 0
x^2 - 28x + 121 = 0
Use the quadratic formula: I got
x ~ 5.34
x ~ 22.66
:
;
7. Calculate the modulus. When necessary, round to the tenths place.
5-3i
Find the absolute of a complex number:
= = = = 5.8

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What are the factors of rx-rsy?
= r*(x - sy)

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two airports, one in california and one in new york, are 3000 miles apart. a plane leaving california is traveling to new york at 200 miles per hour. another leaving new york is traveling to california at 250 miles per hour. when will the two planes pass each other?
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3000/450 = 6 2/3 hour
= 6 hours 40 minutes after takeoff, unless they stop to refuel before that.
Planes that slow seldom have 6 hours of endurance.

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Anything raised to the zero power (except 0 itself) is defined to be 1. You can see it because

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-x^2-2x=5
> x^2+2x+5=0
> x^2+2x+1+4=0
> (x+1)^2=-4
>(x+1)^2=4i^2
So x+1 = +-2i
x= -1+-2i

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(6+3i)-(10+4i)

6+3i-10-4i

(6-10)+(3i-4i)

-4-i


So (6+3i)-(10+4i) = -4-i

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-9i(10i+3)

-9i(10i)-9i(3)

-90i^2-27i

-90(-1)-27i

90-27i

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Determine the derivative of :
f(x)= log x^2
-----------------

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f(x) = ln(x^2)/ln(10)
f'(x) = ((1/x^2)*2x)/ln(10)
f'(x) =

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I need the derivative of : y = sqrt(x^2) + 1 * sqrt(x^2) - 1
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I saw this one yesterday.
It still makes no sense.
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y = sqrt(x^2) + 1 * sqrt(x^2) - 1
y = x + 1*x - 1
y = 2x - 1
y' = 2

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Determine the derivative of :
y = (2x + 3) / (3x + 4)
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y =
y' =
y' =
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It can be rearranged in several forms.
y' =
y' =

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I need the derivative of : y = e^x - e^-x
The answer is supposed to be y' = e^x + e^-x
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The derivative of dx
d(e^x) = e^x
d(e^-x) = -1e^-x

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I need the derivative of : y = radical((x)^2) + 1 * radical((x)^2) - 1
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Something's not right.

The answer is supposed to be y' = 2(x^3)((x^4-1)^-1/2)