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Chikis

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  1. That is an error. I mean to write ₦2000 and not ₦20 000. Please pardon me for that error. 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. 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. 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. 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. No I can't. Could you explain why? Or should I take it that the rule is to use the factors of the constant term.
  6. 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?
  7. [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?
  8. 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? How do you mean? Give a particular example.
  9. What does the r and f stand for? Or what does [math]f( r)[/math] stand for?
  10. What is 'r' and what is [math]f( r)[/math]
  11. It says that [math]x-r[/math] is a factor of a polynomial [math]f( x)[/math] if [math]f( r)=0[/math]
  12. I want to use the factor theorem to find the factors [math]2x^3+x^2-13x+6[/math] How do I do that?
  13. [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.
  14. 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?
  15. I don't know why you don't have to agree. When you have[math]-\left(\frac{a-n}{a+n}\right)[/math], remembes that the minus sign multiplies every term at the numerator to give back your [math]\frac{-a+n}{a+n}[/math] So what is the point here?
  16. Multiplying back? How?
  17. This is the same as[math]-\left(\frac{a+n}{a+n}\right)[/math] Agreed?
  18. Which minus sign are you talking about here?
  19. 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.
  20. 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?
  21. If I multiply [math]\frac{a-n}{a+n}[/math] by minus sign, do I really get [math]-1[/math]
  22. So nothing can be done again?
  23. 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?
  24. 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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