Chikis

On 10/8/2021 at 10:04 AM, PeterBushMan said:

Binary is a base-2 number system that uses two states 0 and 1 to represent a number.

 

Does it mean it only use 2 digits.

 

 

 

Yes of course.  That's 0 and 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?

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

 

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.

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?

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.

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?

[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?

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.

What does the r and f stand for? Or what does [math]f( r)[/math] stand for?

Number 3 is not correct.

What is 'r' and what is [math]f( r)[/math]

It says that [math]x-r[/math] is a factor of a polynomial [math]f( x)[/math] if [math]f( r)=0[/math]

I want to use the factor theorem to find the factors [math]2x^3+x^2-13x+6[/math] How do I do that?

[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.

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?

No, I do not agree.

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?

Multiplying back? How?

So you get[math]-1 \frac{a-n}{a+n} = \frac{-a +n}{a+n}[/math]

This is the same as

[math]-\left(\frac{a+n}{a+n}\right)[/math] Agreed?

You have a minus sign wrong.

Which minus sign are you talking about here?

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.

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?

If I multiply [math]\frac{a-n}{a+n}[/math] by minus sign, do I really get [math]-1[/math]

So nothing can be done again?

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?