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Chikis

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Posts posted by Chikis

  1. Quote

    10% of 20,000 is not 20,000.

    That is an error. I mean  to write ₦2000 and not ₦20 000. Please pardon me for that error.

     

    On 4/15/2019 at 1:09 AM, HallsofIvy said:

    You include 18000 N for "dependent relatives" but I see nothing that says this man HAS any "dependent relatives"! 

    It there in the problem. I copied the the original words in the post i and put the concerned section in bold. You can see it below.

    A man's income is N 180 000 per anum. He is allowed a tax-free income in the following: personal allowance N 72 000, child allowance N 4 500 for each child under 18 years for a maximum of 4 children, dependent relatives allowance to a maximum of N 18 000. Calculate the amount of tax the man will pay if has 5 children under 18 years and the taxable income is 10% for the first N 20 000, 15 % for over N 20 000 and up to N 40 000, 20%  for over 40 000 and up to N 60 000, and 25% for over  ₦60 000.

    Have you seen it?

  2. A man's income is N 180 000 per anum. He is allowed a tax-free income in the following: personal allowance N 72 000, child allowance N 4 500 for each child under 18 years for a maximum of 4 children, dependent relatives allowance to a maximum of N 18 000. Calculate the amount of tax the man will pay if has 5 children under 18 years and the taxable income is 10% for the first N 20 000, 15 % for over N 20 000 and up to N 40 000, 20%  for over 40 000 and up to N 60 000, and 25% for over N 60 000. Here is my work:
    Man's income = N 180 000
    Personal allowance =N 72 000
    Child allowance
    (4 children maximum) = N 4 x N 4 500 = N 18 000
    Dependent relatives allowances = N 18 000

    Total tax free allowances:
    N 72 000 + N 18 000 + N 18 000 = N 108 000
    Taxable income N 180 000 - N 108 000
    = N 72 000
    To calculate the tax on the taxable income:
    N 72 000 = N 20 000 + N 40 000 + N 12 000
    Tax rate on the first N 20 000 is 10%  = N 20 000
    Tax rate over N 20 000 up to N 40 000 is 15% = N 6000

    My problem now is what do I do with the remaining N 12 000 since it is less than N 40 000 and not even over N 65 000? Somebody help me please

     

  3. When it is equal I agree... just that is not true for all a and n.

    And I say give me the value of of a and n, in which it is not true.

    All I can say is that [math]\frac{-a+n}{a+n}[/math] and [math]-\left(\frac{a+n}{a+n}\right)[/math] are the same and diffrent in some cases.

    Nope still completely wrong put 2 and 1 into your first expression and you get a different answer! The problem is that you move from a numerator of (a-n) to a numerator of (a+n) - what simple multiplication does that? Multiplying by -1 would change (a-n) to (-a+n). When moving a minus sign from applying to the whole fraction to just the top OR the bottom - you do it like studiot has explained; you need to read through his stuff again [latex]- \left( \frac{(a - n)}{(a + n)} \right) = \left( \frac{-(a - n)}{(a + n)} \right) = \left( \frac{(-a + n)}{(a + n)} \right)[/latex]

    The first expression is factored to get the second expression. True or force? I will only say that the expression are the same and different.
  4. This is not true.

     

    The minus sign as you write it is 'attached to' a and not (a+n). There is no cancellation here.

     

     

    Try it for yourself carefully...

     

    Let us choose a = 2 n =1.

     

    Then your opening expression is equal to 1/3, which is not 1 or -1. This is enough to show you are wrong.

    Okay, what about when [math]-\frac{a-n}{a+n}[/math]is equal to [math]-\left(\frac{a+n}{a+n}\right)[/math]. The fraction becomes -1. Am sure you agree to this one.

    Okay, coming to when a = 2 and n = 1, if we put the values in the expression, [math]-\left(\frac{a+n}{a+n}\right)[/math], we have [math]-\left(\frac{2+1}{2+1}\right)[/math] [math]=-\left(\frac{3}{3}\right)[/math]. This gives -(1) = -1. What do you have to say about this one?

  5. When [math]x=2[/math]

    [math]\rightarrow[/math]

    [math]2(2)^3+(2)^2-13(2)+6[/math]

    [math]\rightarrow[/math]

    [math]16+4-26+6=0[/math]

    [math]\rightarrow[/math][math]26-26=0[/math]

    [math]\therefore[/math] [math]x-2[/math] is a factor of

    [math]2x^3+x^2-13x+6[/math]

    When [math]x=-3[/math]

    [math]2(-3)^3+(-3)^2-13(-3)+6[/math]

    [math]=2(-27)+9+39+6[/math]

    [math]=-54+48+6[/math]

    [math]=-54+54=0[/math]

    [math]\therefore[/math][math]x+3[/math] is a factor of [math]2x^3+x^2-13x+6[/math] But what is the easiest method for finding the values of x? Or would one be always doing trial and error method?

  6. [math]\frac{3(x+y)(x-y)}{y^2-2xy-3x}[/math]

    Well the solution to the problem above is [math]\frac{(y-x)}{(x-\frac{y}{3})}[/math]. The issue now is that I cannot give any justification to the solution. Any help here?

  7. We are just complicating issue by making this topic long. What I understand and see here is that when you have [math]\frac{a-n}{a+n}[/math] and you multiply the expression by [math]-1[/math],

    it becomes

    [math]\frac{-a+n}{a+n}[/math] The numerator and denominator cancels out to give [math]-1[/math] What are we still aguing here?

    But this is not going to be -1 for all values of a and n. It is if n =0 and a not equal to zero.

    How do you mean? Give a particular example.

  8. [math]\frac{3}{x+1}+\frac{4}{x+5}[/math]

    [math]=\frac{3(x+5)+4(x+1)}{(x+1)(x+5)}[/math]

    [math]=\frac{3x+15+4x+4}{(x+1)(x+5)}[/math]

    [math]=\frac{7x+19}{(x+1)(x+5)}[/math] If the above is what you meant, I have done it and seen that it came out correctly. Thank you.

  9. In the difference of two squares...Expand out (x+y)(x+y) and then (x+y)(x-y) you see that the minus sign makes a big difference.

    Oh! Sorry! I made a mistake there. I intended writing [math]\frac{3(x+y)(x-y)}{y^2-2x-3x}[/math] before I typed that trash. Now that the mistake has been corrected, can the denominator be factored out completely to see the terms that could cancel out?

  10. My reason is this I wanted to simplfy

    [math]\frac{a-n}{a+n}[/math] completely. Looking at the numerator and denominator, these are the same terms which would have cancelled out if not for the opposite signs.

    Knowing that [math](-)(-)=+[/math], I decided to multiply the numerator by minus sign to make the simplification possible.

  11. Just take look at this expression, am required to reduce it to its lowest term.

    [math]\frac{3(x^2-y^2)}{y^2-2xy-3x}[/math]

    Factoring the numerator, we get

    [math]\frac{3(x+y)(x+y)}{y^2-2xy-3x)}[/math], but doing a perfect factoring at the denominator to see how terms could cancel out becomes a major problem. So how does one go about this?

  12. Can [math]y^2-2xy-3x[/math] be factored?

    I tried to do it:

    [math]y^2-(2y-3)x[/math]

    [math]\rightarrow[/math][math]y(y-3)x-1(y-3)x[/math] and that finally gave me

    [math](y-3)x(1+y)[/math] Is this correct?

  13. Well, A and B are constants, but x is a variable - so we can vary it as we like. Can you think of any particular value(s) of x that might be of help in solving this equation?

    Based on the examples I have seen so far, to find A we take that [math]x=-1[/math]

    So [math]7x-1+19=A(-1+5)+B(-1+1)[/math]

    [math]\rightarrow[/math] [math]12=4A+0[/math]

    [math]\rightarrow[/math][math]A=3[/math]

    To find B, we take that [math]x=-5[/math]

    So [math]7x-5+19=A(-5+5)+B(-5+1)[/math]

    [math]\rightarrow[/math] [math]-35+19=0-4B[/math]

    [math]\rightarrow[/math] [math]-16=-4B[/math]

    [math]\rightarrow[/math] [math]B=+4[/math]

    Thus,

    [math]\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}[/math]

    Is my work correct?

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