• Setting Up a Ratio
    June 25, 2017
  • Complex Numbers – Addition and Subtraction
    June 6, 2017

    Complex Numbers

    Complex Numbers
    Addition and Subtraction

    Yesterday we began discussing complex numbers, and I gave you a problem to try involving raising i to a power. You can see that post here: Complex Numbers – Examples and Powers of i

    Today I will go over how to add and subtract complex numbers. But first I will provide a solution to yesterday’s problem. Here is the problem one more time, followed by a solution:

     

    Example: Compute i53.

    Solution: When we divide 53 be 4 we get a remainder of 1. So i53 i= i.

    Notes: (1) For a review of how to compute remainders, see the following post: Remainders

    (2) This computation can also be done quickly in your calculator, but be careful. Your calculator may sometimes “disguise” the number 0 with a tiny number in scientific notation. For example, when we type  i^ 53 ENTER into out TI-84, we get an output of –3E–13 + i. The expression –3E–13 represents a tiny number in scientific notation which is essentially 0. So this should be read as 0 + i = i.

     

    Addition and Subtraction

    We add two complex numbers simply by adding their real parts, and then adding their imaginary parts.

    (a + bi) + (c + di) = (a + c) + (b + d)i

    Example: The sum of the complex numbers 2 – 3i and –5 + 6i is:

    (2 – 3i) + (–5 + 6i) = (2 – 5) + (–3 + 6)i = –3 + 3i

    Subtraction is similar.

    (a + bi) – (c + di) = (a – c) + (b – d)i

    Example 5: When we subtract 2 – 3i from –5 + 6i we get what complex number?

    I will provide a solution to this question tomorrow. Meanwhile, please post your attempted solutions in the comments below.

    For those of you that prefer videos…

     

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  • Complex Numbers – Examples and Powers of i
    June 5, 2017

    Complex Numbers

    Complex Numbers

    A complex number can be written in the form abi where and are real numbers and i2 = –1.

    Examples: The following are examples of complex numbers.

    2 + 3i

    3/2 + (–2i) = 3/2 – 2i

     π + 2.6i

    0 + 5i = 5i                    This is called a pure imaginary number.

    17 + 0i = 17                 This is called a real number.

    0 + 0i = 0                     This is zero.

    Powers of i

    By definition, when we raise any number to the 0 power, we get 1. So in particular, we have i0 = 1.

    Similarly, when we raise any number to the power of 1, we just get that number. So i1 = i.

    By the definition of i, we have i2 = –1.

    i3 = (i2)(i) = (–1)i = –i.

    i4 = (i2)(i2) = (–1)(–1) = 1.

    i5 = (i4)(i) = (1)(i)i.

    Notice that the pattern begins to repeat.

    Starting with i0 = 1, we have

    i0 = 1               i1 = i              i2 = –1             i3 = –i

    i4 = 1               i5 = i              i6 = –1             i7 = –i

    i8 = 1               i9 = i              i10 = –1             i11 = –i

    In other words, when we raise i to a nonnegative integer, there are only four possible answers:

    1, i, –1, or –i

    To decide which of these values is correct, we can find the remainder upon dividing the exponent by 4.

    Example: Compute i53.

    I will provide a solution to this question tomorrow. Meanwhile, please post your attempted solutions in the comments below.

    For those of you that prefer videos…

     

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  • Studying Effectively
    June 4, 2017
  • 28 SAT Math Lessons – Beginner Course – Half Price
    June 2, 2017

    28 SAT Lessons Beginner Front Cover Medium

    28 SAT Math Lessons – Beginner – Half Price

    I’ve just updated the cover of the third book in my 28 SAT Math Lessons series. The book is currently available on Amazon for half of the regular price. Click the following link to get to the book’s Amazon page:

    28 SAT MATH LESSONS – BEGINNER

  • Inverse Variation Question 4 with Solution
    June 1, 2017

    Inverse Variation Question 4 with Solution

    Last week, I went over inverse variation, and I gave you four inverse variation problems to try on your own. You can see that post here: Inverse Variation

    Today I would like to give a solution to the last of those four problems. You can see a solution to the first three problems here: Q1  Q2  Q3

    Example: Suppose that z varies directly as x2 and inversely as y3. If = 9 when = 3 and = 2, what is y when = 4.5 and = 6 ?

    Try to solve the problem yourself before checking the solution below.

    SolutionWe are given that z =  kx2/y3   for some constant k. Since z = 9 when x = 3 and y = 2, we have 9 =  k(3)2/23   =  9k/8. So k = 8, and z =  8x2/y3 . We now substitute z = 4.5 and x = 6 to get 4.5 =  8(6)2/y3 . So y3 =  8(36)/4.5  = 64, and therefore y = 4, choice D.

     

    More Problems with Explanations

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  • Inverse Variation Question 3 with Solution
    May 31, 2017

    Inverse Variation Question 3 with Solution

    Last week, I went over inverse variation, and I gave you four inverse variation problems to try on your own. You can see that post here: Inverse Variation

    Today I would like to give a solution to the third of those four problems. You can see a solution to the first two problems here: Q1  Q2

    Example: If x ≠ 0 and x is directly proportional to y, which of the following is inversely proportional to  1/y2 ?

    A) x2
    B) x
    C) 1/x
    D) 1/x2

    Try to solve the problem yourself before checking the solution below.

    SolutionSince x is directly proportional to y, x = ky for some constant k.

    Taking the reciprocal of each side of this equation gives us 1/x = 1/ky.

    Cross multiplying yields kyx

    So kx(1/y). Squaring each side of this last equation gives k= x2(1/y2).

    Since k2 is a constant, x2 is inversely proportional to 1/y2. So the answer is choice A.

     

    More Problems with Explanations

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  • Inverse Variation Question 2 with Solution
    May 30, 2017

    Inverse Variation Question 2 with Solution

    Last week, I went over inverse variation, and I gave you four inverse variation problems to try on your own. You can see that post here: Inverse Variation

    Today I would like to give a solution to the second of those four problems. You can see a solution to the first problem here: Inverse Variation Q1

    Example: If x varies inversely as y2, and x is 3 when y is 5, then what is x when y is 3 ?

    Try to solve the problem yourself before checking the solution below.

    SolutionWe are given that y = 5 when x = 3. Since  x varies inversely as y2, we have that xyis a constant. So we get the following equation: (3)(25) = x(9). So 75 = 9x, and therefore x = 75/9 = 25/3.

     

    More Problems with Explanations

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  • Inverse Variation Question 1 with Solutions
    May 29, 2017

    Inverse Variation Question 1 with Solutions

    Last week, I went over inverse variation, and I gave you four inverse variation problems to try on your own. You can see that post here: Inverse Variation

    Today I would like to give a solution to the first of those four problems. I will give solutions to the other three problems throughout this week.

    Example: If y = k/x and y = 9 when x = 14, then what is y when x = 6 ?

    Try to solve the problem yourself before checking the solutions below.

    Solutions

    (1) We are given that y = 9 when x = 14, so that 9 = k/14, or k = 126. Thus, y = 126/x. When x = 6, we have y = 126/6 = 21.

    (2) Since y varies directly as x,  xy  is a constant. So, we get the following equation:  (14)(9) = 6y. So, 126 = 6y, and y = 126/6 = 21.

    (3) (14)(9)/6 = 21.

     

    More Problems with Explanations

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  • Inverse Variation
    May 27, 2017

    inverse variation graph

    Inverse Variation

    Last week, I gave you several posts on direct variation. You can see those posts by clicking the following links: Direct Variation  Q1  Q2  Q3

    Today I would like to talk about inverse variation.

    The following are all equivalent ways of saying the same thing:

    (1) y varies inversely as x. 
    (2) y is inversely proportional to x.
    (3) = k/x for some constant k.
    (4)  xy  is constant.

    The following is a consequence of (1), (2), (3), or (4)

    (5) the graph of = f(x) is a hyperbola

    Note: (5) is not equivalent to (1), (2), (3), or (4).

    Example: In the equation y = 12/x, y varies inversely as x. Here is a partial table of values for this equation.

    Note that we can tell that this table represents an inverse relationship between x and y because (1)(12) = (2)(6) = (3)(4) = (4)(3) = 12. Here the constant of variation is 12.

    Here is a graph of the equation. On the left, you can see the full graph. On the right, we have a close-up in the first quadrant.

    The various equivalent definitions of direct variation lead to several different ways to solve problems.

    Example: If y = k/x and y = 8 when x = 3, then what is y when x = 6 ?

    Try to solve the problem yourself before checking the solutions below.

    Solutions

    (1) We are given that y = 8 when x = 3, so that 8 = k/3, or k = 24. Thus, y = 24/x. When x = 6, we have y = 24/6 = 4.

    (2) Since y varies directly as x,  xy  is a constant. So, we get the following equation:  (3)(8) = 6y. So, 24 = 6y, and y = 24/6 = 4.

    (3) (8)(3)/6 = 4.

     

    Here are a few more problems for you to try. I will provide solutions to these over the next few days.

    1. If y = k/x and y = 9 when x = 14, then what is y when x = 6?

    2. If x varies inversely as y2, and x is 3 when y is 5, then what is x when y is 3

    3. If x ≠ 0 and x is directly proportional to y, which of the following is inversely proportional to  1/y2 ?

    A) x2
    B) x
    C) 1/x
    D) 1/x2

    4. Suppose that z varies directly as x2 and inversely as y3. If = 9 when = 3 and = 2, what is y when = 4.5 and = 6 ?

     

    More Problems with Explanations

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  • 28 SAT Math Lessons – Intermediate Course – Half Price
    May 24, 2017