Maths Game (2 Viewers)

[pwoh]

Same here, I always get the easy questions wrong by doing something stupid.

Someone else provide questions please?

 

[Somnolence]

Simplify (Non-Calculator)

321*23*9
________
642*69*5

This is a similar question to one we had in our trials. To some it may appear difficult, but once you look at it closely, it is really easy.

 

[pwoh]

Simplify (Non-Calculator)

321*23*9
________
642*69*5

This is a similar question to one we had in our trials. To some it may appear difficult, but once you look at it closely, it is really easy.

1*23*9
________
2*69*5

1*23*3
________
2*23*5

3/10

 

[Air Jordan]

[log(base 2)x] + [log(base 2) (x²-4)] - [log(base2)(x+2)] = 3

 

[suzlee]

[log(base 2)x] + [log(base 2) (x²-4)] - [log(base2)(x+2)] = 3

logarithm is an optional topic. It's not gonna be in the SC............................

 

[suzlee]

o_o; Is log a part of one of the optional topics?

We haven't done it yet. Someone else answer...

I haven't done it either. Yes it is optional, along with polynomials, circle geometry, etc

 

[pwoh]

logarithm is an optional topic. It's not gonna be in the SC............................

I wonder if there's someone who can answer that anyway.

Jenni bought 30CDs and DVDs for a total of $855. Each CD cost $20 and each DVD cost $35. How many of each did she buy?

 

[suzlee]

I wonder if there's someone who can answer that anyway.

Jenni bought 30CDs and DVDs for a total of $855. Each CD cost $20 and each DVD cost $35. How many of each did she buy?

yay finally a SC question


number of CDs = x

855 = 20x + 35(30-x)
855 = 20x + 1050 - 35x
195 = 15x
13 = x


She bought 13 CDs and 17 DVDs





oh yeah I'll post a question:
Four years ago, I was twice as old as I was 14 years ago. What is my age?

 

[Miss Sunshine]

age now= x

x-4=2(x-14)
x-4=2x-28
x+24=2x
x=24

Therefore 4 years ago I was 20

14 years ago i was 10

So double 10 is 20

Your age now is 24.

 

[hn92]

taken from baulko year 10 trial, quite a simple but awesome question.

In the Diagram below ABCD and DEFG are squares,

show that CF² + BE² = 3(AD² + DE²)

http://img171.imageshack.us/my.php?image=squarzdb9.png

AWESOME??

awesome is not equal to a maths equation lol

theres some 'awesome maths' for you.

 

[Kabuzo01]

I cant be bothered updating points yet... Too busy with the school thread =P

 

[Lukybear]

[log(base 2)x] + [log(base 2) (x²-4)] - [log(base2)(x+2)] = 3

Might be wrong but:
log(base2)((x(x^2-4))/x+2))=3
8=x(x^2-4)/x+2
(factorise)
x(x-2)(x+2)/x+2=8
x^2-2x-8=0
factorise
x=4 or -2

is that rite?

 

[Anonymou$]

True. But there are exceptions to every rule.

Is there an exception to that rule?

 

[DALiE. (:]

Is there an exception to that rule?

Possibly.

 

[lyounamu]

Might be wrong but:
log(base2)((x(x^2-4))/x+2))=3
8=x(x^2-4)/x+2
(factorise)
x(x-2)(x+2)/x+2=8
x^2-2x-8=0
factorise
x=4 or -2

is that rite?

Haven't read your whole working out but you cannot have x=-2.

 

[xXmuffin0manXx]

Might be wrong but:
log(base2)((x(x^2-4))/x+2))=3
8=x(x^2-4)/x+2
(factorise)
x(x-2)(x+2)/x+2=8
x^2-2x-8=0
factorise
x=4 or -2

is that rite?

no

3

 

[lyounamu]

no

3, -1

You are wrong. I just did the question.

I am sure it's x=4.

 

[xXmuffin0manXx]

You are wrong. I just did the question.

I am sure it's x=4.

Oh. Okay then



where did the 8 come from?
cant find out how

 

[Air Jordan]

Yeah. x=4

 

[lyounamu]

Oh. Okay then



where did the 8 come from?
cant find out how

Would you like to elaborate on that? I don't understand what you mean.