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#1 2015-11-15 09:19:46

evene
Member
Registered: 2015-10-18
Posts: 272

Need help!

Lots of questions, but the more the merrier! (I guess)


(1) Simplify


(2) Divide the face of a clock into 3 parts with 2 lines so that the sum of the numbers in the 3 parts are equal


(3) How many squares are there in a 12 by 12 board?


(4) There are 25 people in a room. 10 people are wearing socks and 18 people are wearing shoes, 7 people are wearing both. How many people are in bare feet?




Thank you for your patience!

Last edited by evene (2015-11-15 09:20:01)

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#2 2015-11-15 20:20:46

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Need help!

hi evene

Merriness is what we're here for!  smile

Q3.  This looks like an extension of the "How many squares on a chess board?" question.  There's one square that is 12 x 12; four that are 11 x 11; nine that are 10 x 10 and so on.  There is a formula for summing square numbers but there's not so many that you cannot just add them up.

Q4.  I would draw a Venn diagram.  (1) Draw a rectangle to contain the 25 people.  Draw two overlapping circles for socks and for shoes.  Write 7 in the overlapping part.

Subtract 7 from 10 and write the result in the socks but not shoes area.  Work out a similar number for shoes but not socks.  You have now entered three of the four numbers for the regions.  If you subtract their total (only counting 7 once!) you will have the number for the 'outside the circles' area.

Q2.  The sum 12 + 11 + 10 + ....+ 1 = 78.  So each part must add to 78/3 = 26.  12 must be in a group along with numbers adjacent to 12 that add to 26.  I only found one way of doing this.  So splitting the remaining numbers into two more lots of 26 was then easy.

Q1.  You can make use of "the difference of two squares" again and again.

and so on.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2015-11-16 12:39:29

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Thanks, you make them look so easy. I feel like I'm overcomplicating them. Not sure if that's a good thing or bad thing

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#4 2015-11-17 14:54:57

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

For (3), won't you also have to add 144 to whatever number you get adding 1+4+9... because there are also 144 small 1x1 squares in the board too! Just a thought. And, if you know the formula, please tell me. My hands are aching from writing so much and adding so many things together! D:

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#5 2015-11-17 19:14:36

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Need help!

Q3)

There are

The general formula is:


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#6 2015-11-18 01:38:47

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Ok, thanks

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#7 2015-11-18 05:52:07

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Need help!

The formula is for an n x n square.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#8 2015-11-21 14:53:04

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

A new problem, this one should be easy compared to the other ones; I'm just a bit flustered on where to start.

The solution of a system of equations is (-3, 2). One equation in the system is

. Find a second equation for the system. Explain how you derived this equation.

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#9 2015-11-21 16:17:19

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Need help!

Hi;

There are zillions of solutions here, one is x + y = - 1. Do you see why?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#10 2015-11-22 02:06:36

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Yeah, I understand why there are an

solutions here. After playing around with it, I have one:
Although is there a solid method to get such equation?

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#11 2015-11-22 02:13:32

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Do you guys here know any mental math tricks? Like how to multiply any number by 11 mentally, 12, 5, etc..

I know the 11 trick and the squaring any number ending in 5 trick. But that's about it

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#12 2015-11-22 04:12:28

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Need help!

hi evene,

Is there a method ?  (-3,2) is pointing me towards the equation of a straight line http://www.mathsisfun.com/algebra/syste … tions.html

The given line goes through that point and so will any other equation for which this is the solution.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#13 2015-11-22 05:37:01

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Need help!

Confucius wrote:

Never use a cannon to kill a mosquito.

Although is there a solid method to get such equation?

The general form for a straight line is ax + by = d. Since you are given that this line goes through (-3,2) that becomes 2b - 3a = d. Now just substitute any number for a and b.

Example 1:

ax + by = d

Pick a = 3 and b = 5

3x + 5y = d

We know that x = - 3 and y = 2.

5(2) + 3(-3) = d =1 so our equation is 3x + 5y = 1

Example 2:

ax + by = d

Pick a = 9 and b = 11

9x + 11y = d

We know that x = - 3 and y = 2.

9(-3) + 11(2) = d = -5

9x +11y = -5

You can continue this process...


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#14 2015-11-22 05:41:48

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Need help!

hi bobbym,

Are you sure Confucius said that?  I thought the cannon was invented a little after Confucius.  smile

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#15 2015-11-22 05:49:06

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Need help!

Hi;

I think it was one of his descendants, Johann Confucius.

The legendary C. Greathouse used that quote on me in response to one of my answers, so I keep it intact. It is a reminder to not use the heavy machinery unless forced to.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#16 2015-11-22 08:00:48

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

I'm pretty sure that Confucius actually did say that saying

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#17 2015-11-22 08:15:59

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Need help (now) with a little bit of Quadratic stuff. And please show work! Need to understand how you guys solved it! smile


(1) If both -3 and

are solutions to the equation
, where a, b and c are integers, what is a possible value of 2a+b-c?

(2) Given an equation

, if the sum of two solutions to this equation is
and the product of the two solutions is -4;

(a) What are the values of b and c

(b) What are the solutions to this equation?

(3) Use the Synthetic Division Theorem to do each of the following dividion.

(a)

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#18 2015-11-22 09:23:22

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Need help!

hi evene,

Q1. As you know that x = -3 or 2/3 you can say that (x+3) and (3x-2) are factors.  Multiply them and you'll get a quadratic.  You can pick out values for a, b and c from this.

[pedantic note (ignore if you wish):  So we end up with numbers to fit ax^2 + bx + c = 0.  But those solutions will also fit akx^2 + bkx + ck = 0 for any integer k.  So 2a + b - c could take the value k(2a + b - c).  In other words any number in a certain times table is a possible answer.  roflol

Q2.  I would call the solutions alpha and beta.  So

So you can write

Make alpha the subject of one of these and substitute into the other.  You get a quadratic which will tell you b and c.  And you can then solve it.  (note: If instead, you seek to eliminate beta you'll end up with the same quadratic so it doesn't matter which way you do it.)

Q3.  It takes ages to format this nicely for a post so I hope you know roughly what this method involves.

step 1.  To get -x^4 you need to multiply the divisor by -2x^3.  Subtracting gives -4x^3.

step 2.  To get that you need to multiply the divisor by +4x^2.  Subtracting gives 11x^2.

etc etc.

I made the final remainder 52.  Hope that is enough.   

The method is nicely demonstrated here: http://www.mathsisfun.com/algebra/polyn … -long.html

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#19 2015-11-22 09:26:51

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Thanks Bob! This was really helpful. I refer the website for Q3

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#20 2015-11-22 09:28:10

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Need help!

I have just made a few edits!!! to the above.  Please check it again.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#21 2015-11-22 09:30:59

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Ah..

Looking online revealed this method for question 2:

let s and t be solutions to the quadratic

s+t=1/2 and st=-4

Replacing s+t with -b/a and st with c/a gives:

-b/a=1/2 and c/a=-4
Plugging in a=2 gives b and c, which you plug into the original equation

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#22 2015-11-22 09:36:22

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

Need help with this problem. I know how to do this one, but I am getting a really weird number and I am pretty superstitious about this:

Use substitution to answer the question:

(1)


Let

I don't think this is factor-able and when I use the formula, I get

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#23 2015-11-22 09:46:51

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Need help!

The s and t method is much the same as what I suggested, so I'm happy with that.

It isn't factorable and that solution is right:

You can get x from those two values but it's tricky.  How about doing it like this:

times throughout by x^2

Then you can use the quadratic formula on that for immediate solutions.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#24 2015-11-22 09:50:14

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Need help!

That would work, but I am forced to use the substitution method that I said on the last post, so... "yay" me!

Alright well, better start working... hmm

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#25 2015-11-22 09:57:01

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Need help!

Hi;

You got the right answer but you solved for y. You want x, so y = 1/x and x = 1 / y. Can you solve now?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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