Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

Pages: 1 … 12 13 14 15 16

#326 Re: Help Me ! » Can sqrt x be a factor? » 2023-11-11 00:10:17

paulb203

Hi, Bob.

“...as root 9 = 3, you have already listed it.  If the answer to a question was 1/2 then 0.5, sin(30), and 16/32 would all be equivalent ways of saying the same thing.”

Makes sense. Thanks.

“In all the books, lessons and exam papers I have encountered, where 'factor' of an integer has arisen, it has always meant positive whole number(s) that divide exactly.”
Interesting. I’m discussing this in another thread with Jai Ganesh (“Factors. Can factors be negative numbers?”)

#327 Re: Help Me ! » Factors. Can negative numbers be factors? » 2023-11-10 23:55:52

paulb203

Thanks, Jai Ganesh

Interesting. Because the first half dozen or so Google results I got said that factors can be negative.

It does seem right that they would count as factors. We're told that a factor is a number that when multiplied by certain other numbers gives us a particular product (2, when multiplied by 9, gives us the product 18). -2 and -9 seem to fit this criteria.

But I trust your good self, and this site generally, so I'm now doubting those Google results, and my own intution.

Who, or what body, is the ulitmate arbiter of these things? Is there an ultimate arbiter of these things?

#328 Help Me ! » Can sqrt x be a factor? » 2023-11-10 23:13:03

paulb203
Replies: 2

Example;

Is sqrt of 9 a factor of 9?

sqrt 9 times sqrt 9 = 9

If so, should we list sqrt when listing the factors of 9?

Should the list be 1,3,9,sqrt 9 (and their negative counterparts, if those count as factors)?

#329 Help Me ! » Factors. Can negative numbers be factors? » 2023-11-10 23:03:32

paulb203
Replies: 7

A question on Maths Genie;

Write down all the factors of 14.

My answer; -1,-2,-7,-14,1,2,7,14

Their answer; 1,2,7,14

Which is correct

#330 Maths Is Fun - Suggestions and Comments » I suggest having a like button » 2023-11-01 05:23:14

paulb203
Replies: 2

It's a good, brief way of acknowledging a post when commenting on it isn't necessary, I think.

#331 Re: Help Me ! » Surds. Division Rules. » 2023-10-31 03:47:13

paulb203

Thanks, guys.

#332 Re: Help Me ! » Surds. Division Rules. » 2023-10-30 00:02:49

paulb203

Thanks, Bob.

I'm getting the impression that BODMAS is for primary and secondary school maths, but at the more advanced level of degree and post graduate it doesn't really apply as maths problems at that level are set out unambiguously, e.g, instead of the division symbol you would have a slash; you wouldn't have the times symbol (x) you would have brackets instead, or the multiplication would be implied by juxtaposition (e.g, a number directly in front of brackets).

I spoke to a few others online about this, each of whom got annoyed at the mere question, as they were sick of what they regarded as, 'all this BODMAS' nonsense.

It made me think, Who sets the agenda? And why don't the early years educators get together with the more advanced people and create an objective standard?

#333 Re: Help Me ! » Surds. Division Rules. » 2023-10-29 23:29:25

paulb203
Jai Ganesh wrote:

Hi,


BODMAS: Brackets Orders Division Multiplication Addition Subtraction.
Brackets make a difference.

Thanks. Why have you added brackets?

#334 Help Me ! » Surds. Division Rules. » 2023-10-29 00:38:44

paulb203
Replies: 9

Why does, 4√64÷2√4
=2√16 ?

I can see the method (4÷2=2, √64÷√4=√16)

But why is it not;

4 times √64 divided by 2 times √4 (as per BODMAS)
i.e, 4 times 8 divided by 2 times 2, which = 32

Nb. When I put it into my calculator that way it gives 32 as the answer
(4*√64÷2*√4)

#335 Help Me ! » x^2-9y^2=0 (where x>0 and Y>0) Work out the ratio x:y » 2023-10-19 07:37:29

paulb203
Replies: 2

x^2-9y^2=0 (where x>0 and Y>0)
Work out the ratio x:y

Not sure where to start with this one.

Take -9y^ from both sides? Not sure if that achieves anything.
x^2 = -9y^2

Square root of both sides?
What’s the square root of –9y^2?

Tough one...

EDIT;

If I take -9y^2 from both sides it should then be:

x^2 = 9y^2

As for the square root of 9y^2..?

That is 3y, yeah?

So if I square root both sides I get:

x=3y?

#336 Re: Help Me ! » Quadratic equations. Solution for x. Is it AND, or, OR? » 2023-10-18 23:05:00

paulb203

Thanks.
So one answer will work, when plugged in to the equation. And one won't work?

#337 Help Me ! » Quadratic equations. Solution for x. Is it AND, or, OR? » 2023-10-18 22:45:46

paulb203
Replies: 4

When we solve for x and get two answers do we state this as, for example, x = -2 OR x=3. Or do we state it as x = -2 AND x=3?

I'm new to quadratic equations, obviously, and I've heard someone say, 'Or', another say, 'And', and yet another just put a comma between the two answers.

#338 Re: Help Me ! » How do I know if I've successfully drawn a cube? » 2023-10-13 05:06:16

paulb203

Thanks, everybody.

Another one that was way more complicated than I thought it would be smile

Got to love maths!

#339 Re: Help Me ! » Ratio. Direct Proportion. Amounts change causing ratio to change, etc. » 2023-10-13 05:00:47

paulb203

I learned a nice way to solve this kind of problem today:


Initial ratio 5:6
J = 5x
K = 6x
Then, after J gains 2:
Ratio becomes 7:8
J = 5x+2
K = 6x
Therefore:
J/K = 7/8
and:
J/K = 5x+2/6x
Therefore:
5x+2/6x = 7/8
Cross multiply to give
42x = 40x+16
2x=16
x=16/2
x=8
5x=40
6x=48
Jane owned 40 initially
Kim owned 48 initially

I know I've ended up with the correct answer, but is the working and the method all correct?

#340 Re: Help Me ! » Solving equations using ratio » 2023-10-13 04:56:41

paulb203

Thanks, amnkb,

I'll need to have a long, hard think about that one smile

#341 Re: Help Me ! » Solving equations using ratio » 2023-10-13 04:55:38

paulb203

Just tackled the middle part, Bob. I wasn’t sure why
7xy divided by y^2 = 7 times x/y
I think I’ve got it now:
7xy/y^2 = 7x/y (cancel out one of the y’s on the top, and one of the y’s on the bottom)
And then 7x/y is the same as 7 times x/y (just like in the first part)
Is that right?

#342 Re: Help Me ! » Ratio. Direct Proportion. Amounts change causing ratio to change, etc. » 2023-10-12 10:49:47

paulb203

Thanks, amnkb. I think I've got it. 48 as a common denominator gives, 40/48, and, 42/48. Which is what I got when I did it another way, using trial and error.

amnkb wrote:
paulb203 wrote:

Jane and Kim own some marbles in the ratio 5:6
Jane gains 2 more marbles and the ratio is now 7:8
How many marbles do each of them own initially?

the ratios tell you that there is some factor that divides out of jane/kim
the first ratio simplifies as jane/kim = 5/6
the second ratio simplifies as jane/kim = 7/8
try common denominators
5/6 = 20/24
7/8 = 21/24
not 2 apart
try other common denominators

#343 Re: Help Me ! » Solving equations using ratio » 2023-10-12 06:08:05

paulb203

Thanks, Bob.

This one's a challenge!

I was puzzled at;

6x^2 / y^2 = 6 (x^2/y^2)

But I think I might have understood that with a bit of work. Tell me if I'm right, please:

6 of something / something IS THE SAME AS: 6 of (the first something / the second something)?

Example:

2 of 6 / 4 = 12/4 = 3

and,

2 of (6/4) = 2 of 1.5 = 3

?

As for the rest: I'll need to take it in stages (after 2 paracetamol and a lie down smile)

#344 Re: Help Me ! » Ratio. Direct Proportion. Amounts change causing ratio to change, etc. » 2023-10-12 05:30:30

paulb203

Beautiful. Thanks, Bob.

So,

(x+2).8/7 = 6x/5

8/7(x+2) = 6x/5

(8x+16)/7 = 6x/5

Now what?

Times both sides by 7?

Times both sides by 5?

I’m not sure how to do either.

Try times both sides by 7:

(8x+16) = ?

How to do, 7(6x/5)?

Try: 7/1 x 6x/5 = 42x/5 ?

So, (8x+16)= 42x5 (I hope!)

Times boths sides by 5?

**90x+80 = 42x**

90x+80-42x=0

48x+80=0

48x=(-80)

x=(-80/48)

x= -5/3?

That can’t be right smile

Where have I gone wrong.....?

**EDIT**

Should've been 40x+80=42x
42x-40x = 80
2x=80
x=40

Which is what I got using trial and error; and then again when amnkb helped me do it his ingenious way.

#345 Help Me ! » Solving equations using ratio » 2023-10-12 03:54:20

paulb203
Replies: 5

I'm not even sure where to start with this one.

6x^2=7xy+20y^2 where x>0 and Y>0

Find the ration of x:y

I was thinking, rearrange? Simplify? Factorise? But didn't get very far.

Again, a few hints initially would be appreciated, rather than an explanation wholesale.

#346 Help Me ! » Ratio. Direct Proportion. Amounts change causing ratio to change, etc. » 2023-10-12 03:22:56

paulb203
Replies: 5

Jane and Kim own some marbles in the ratio 5:6
Jane gains 2 more marbles and the ratio is now 7:8
How many marbles do each of them own initially?

I can solve these problems using trial and error with the numbers but I don't know the simplest way of solving them using basic algebra.
Please don't tell me how to do that fully. Just a hint or two would be helpful initially.

Do please tell me the answer though (how many they each own initially), to check with my trial and error answer.

#347 Help Me ! » Ratio. £360 shared between A, B, C and D... » 2023-10-07 09:56:24

paulb203
Replies: 1

Ratio A:B = 2:7

C and D each 1.5 x A

How much does B get?

*

I can get the correct answer. But I still have one question. I'll leave it for now.

How would you go about solving this?

#348 Help Me ! » How do I know if I've successfully drawn a cube? » 2023-10-07 07:20:28

paulb203
Replies: 9

If I draw a square, 2cm by 2cm, then draw another square, also 2cm by 2cm, starting the second from the centre of the first, in the manner children attempt to draw their first cube (and in the manner I still occasionally try to draw a cube and wonder if it’s actually a cube. Then I join the corners of the two squares in the same manner, have I actually drawn a cube?

When I look at the finished drawing it appears to be a cube. But the only measurements I’m certain of are the respective lengths and breadths of the two squares, each 2 by 2.

The four lines used to join the two squares, to give the cube its depth, are less than 2cm, on the page. If it was an actual cube in front of me, an actual 3d solid shape, those ‘depth’ lines would measure 2cm also, obviously. Does it matter that the depth lines are not 2 cm on the page? Is it still a cube?

#349 Re: Help Me ! » Factorising a quadratic expression where the a term is greater than 1 » 2023-10-07 02:27:57

paulb203

Thanks, Bob.

But that level of algebra is too difficult for me.

#350 Re: Help Me ! » Factorising a quadratic expression where the a term is greater than 1 » 2023-10-06 06:09:31

paulb203

@Bob
“Can you show me an example where it makes a real difference?”
I believe it makes a real difference in your example, but I’ll give another in case I’ve not been clear.
4x^2 -19x + 12
1.    Multiply the a term by the c term, i.e, 4x12=48
2.    Choose a factor pair from 48 that will add to give the b term, the middle term, i.e, -19. That pair will be -3 and -16.
3.    Write out your bracket pair but put 4x in each pair instead of 4x in one, and x in the other.
4.     Insert your factor pair, to give:
(4x-3)(4x-16)
5.    Cancel where you can, i.e, (4x-16) cancels to (x-4)
6.    Which gives: (4x-3)(x-4), which is your answer.
7.    Nb. The only extra step is the cancelling which takes less time than fiddling with the factor pair.

Pages: 1 … 12 13 14 15 16

Board footer

Powered by FluxBB