[max25]

(Original post by As_Dust_Dances_)
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1

no it was 10y +6 the perimeter!

[I_am_god_123]

(Original post by As_Dust_Dances_)
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1

total perimeter was 10Y+6

[taylor56]

(Original post by As_Dust_Dances_)
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1

total perimeter was 10y+6

[KoalaKim]

(Original post by As_Dust_Dances_)
The total perimeter was 10y+3 and it had to be between 20 and 54 right?
so 20<10y+3<54
17<10y<51
1.7<y<5.1

10y +6

check out http://www.thestudentroom.co.uk/show....php?t=2002389

unofficial mark scheme

[benex381]

i got -7<x<4 and 1.7<x<5.1 ...

[max25]

(Original post by I_am_god_123)
total perimeter was 10Y+6

20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!! !!

[KoalaKim]

(Original post by max25)
how did you find the last question?

it was ok, pretty sure i got it right

check out

http://www.thestudentroom.co.uk/show....php?t=2002389

[adamjay]

(Original post by As_Dust_Dances_)
For the perimeter question I got 17/10<x<51/10

I thought it was in terms of y?

I think I got something like this...

34 <= 10y + 6 <= 54

When simplified gets:

28 <= 10y <= 48

then:

14/5 <= y <= 24/5

[max25]

(Original post by I_am_god_123)
total perimeter was 10Y+6

eeek apologies, wrong question hehe!were both right

[As_Dust_Dances_]

(Original post by I_am_god_123)
nah it was -7<x<4
but then x must be +ve

for part i) (x+3)(4x)<112 right?

so 4x^2 +12x-112<0

x^2+3x-28

(x+7)(x-4)

You're talking about part 1 right, I was on about part 2..

So yeah it is x<-7 and x>4

[billybobs]

(Original post by As_Dust_Dances_)
for part i) (x+3)(4x)<112 right?

so 4x^2 +12x-112<0

x^2+3x-28

(x+7)(x-4)

You're talking about part 1 right, I was on about part 2..

So yeah it is x<-7 and x>4

But x cannot be negative - you cannot have a negative length as one of the sides was 4x negative dimensions are impossible so it had to be 0>x>4

[Venomilys]

100 ****ing percentaaah!

[KoalaKim]

(Original post by Ilyas)
100 ****ing percentaaah!

check this out...
http://www.thestudentroom.co.uk/show...389&p=37573386

[I_am_god_123]

(Original post by max25)
20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!! !!

yaeh that's correct

[PhysicsMan]

Question 6

Equation of normal is

4x - 6y -29 = 0

[benex381]

would i lose a mark for having it as -4x + 6y + 29?

[taylor56]

(Original post by As_Dust_Dances_)
for part i) (x+3)(4x)<112 right?

so 4x^2 +12x-112<0

x^2+3x-28

(x+7)(x-4)

You're talking about part 1 right, I was on about part 2..

So yeah it is x<-7 and x>4

this is right.

[As_Dust_Dances_]

(Original post by PhysicsMan)
Question 6

Equation of normal is

4x - 6y -29 = 0

This rings a bell has the gradient was like 2/3 or something?

[PhysicsMan]

(Original post by As_Dust_Dances_)
This rings a bell has the gradient was like 2/3 or something?

Yep, the gradient of tangent was -3/2 therefore the gradient of normal is 2/3.

[Mattywooda]

(Original post by max25)
20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!! !!

That can't be correct (I didnt sit the paper though).. You are saying 20 < value > 54... so 20 is bigger than 54? 20 >54: you needed two statements.