 # Chikis

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156

## Posts posted by Chikis

### Calculate the amount of tax

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?

### Calculate the amount of tax

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

### Multiplying agebriac fraction by a minus sign

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 $\frac{-a+n}{a+n}$ and $-\left(\frac{a+n}{a+n}\right)$ 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 $- \left( \frac{(a - n)}{(a + n)} \right) = \left( \frac{-(a - n)}{(a + n)} \right) = \left( \frac{(-a + n)}{(a + n)} \right)$

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.

### Multiplying agebriac fraction by a minus sign

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 $-\frac{a-n}{a+n}$is equal to $-\left(\frac{a+n}{a+n}\right)$. 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, $-\left(\frac{a+n}{a+n}\right)$, we have $-\left(\frac{2+1}{2+1}\right)$ $=-\left(\frac{3}{3}\right)$. This gives -(1) = -1. What do you have to say about this one?

### Factor theorem

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.

### Factor theorem

When $x=2$

$\rightarrow$

$2(2)^3+(2)^2-13(2)+6$

$\rightarrow$

$16+4-26+6=0$

$\rightarrow$$26-26=0$

$\therefore$ $x-2$ is a factor of

$2x^3+x^2-13x+6$

When $x=-3$

$2(-3)^3+(-3)^2-13(-3)+6$

$=2(-27)+9+39+6$

$=-54+48+6$

$=-54+54=0$

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

### Can this be factored?

$\frac{3(x+y)(x-y)}{y^2-2xy-3x}$

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

### Multiplying agebriac fraction by a minus sign

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

it becomes

$\frac{-a+n}{a+n}$ The numerator and denominator cancels out to give $-1$ 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.

### Factor theorem

What does the r and f stand for? Or what does $f( r)$ stand for?

### Multiplying agebriac fraction by a minus sign

Number 3 is not correct.

### Factor theorem

What is 'r' and what is $f( r)$

### Factor theorem

It says that $x-r$ is a factor of a polynomial $f( x)$ if $f( r)=0$

### Factor theorem

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

### Partial Fraction

$\frac{3}{x+1}+\frac{4}{x+5}$

$=\frac{3(x+5)+4(x+1)}{(x+1)(x+5)}$

$=\frac{3x+15+4x+4}{(x+1)(x+5)}$

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

### Can this be factored?

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 $\frac{3(x+y)(x-y)}{y^2-2x-3x}$ 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?

### Multiplying agebriac fraction by a minus sign

No, I do not agree.

I don't know why you don't have to agree. When you have

$-\left(\frac{a-n}{a+n}\right)$, remembes that the minus sign multiplies every term at the numerator to give back your $\frac{-a+n}{a+n}$ So what is the point here?

### Partial Fraction

Multiplying back? How?

### Multiplying agebriac fraction by a minus sign

So you get$-1 \frac{a-n}{a+n} = \frac{-a +n}{a+n}$

This is the same as

$-\left(\frac{a+n}{a+n}\right)$ Agreed?

### Can this be factored?

You have a minus sign wrong.

Which minus sign are you talking about here?

### Multiplying agebriac fraction by a minus sign

My reason is this I wanted to simplfy

$\frac{a-n}{a+n}$ 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 $(-)(-)=+$, I decided to multiply the numerator by minus sign to make the simplification possible.

### Can this be factored?

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

$\frac{3(x^2-y^2)}{y^2-2xy-3x}$

Factoring the numerator, we get

$\frac{3(x+y)(x+y)}{y^2-2xy-3x)}$, 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?

### Multiplying agebriac fraction by a minus sign

If I multiply $\frac{a-n}{a+n}$ by minus sign, do I really get $-1$

### Can this be factored?

So nothing can be done again?

### Can this be factored?

Can $y^2-2xy-3x$ be factored?

I tried to do it:

$y^2-(2y-3)x$

$\rightarrow$$y(y-3)x-1(y-3)x$ and that finally gave me

$(y-3)x(1+y)$ Is this correct?

### Partial Fraction

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 $x=-1$

So $7x-1+19=A(-1+5)+B(-1+1)$

$\rightarrow$ $12=4A+0$

$\rightarrow$$A=3$

To find B, we take that $x=-5$

So $7x-5+19=A(-5+5)+B(-5+1)$

$\rightarrow$ $-35+19=0-4B$

$\rightarrow$ $-16=-4B$

$\rightarrow$ $B=+4$

Thus,

$\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}$

Is my work correct?

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