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The diagram shows a rectangular birthday card which is x cm wide and (x + 8) cm tall. Given that the height of the card is to be at least 50% more than its width, a show that x ≤ 16.
Given also that the area of the front of the card is to be at least 180 cm2 , b find the set of possible values of x.
Could someone help me to answer this question?
(edited 7 years ago)
I would really really really help but I suck at maths and have no idea what you're going on about.
Here's your free reply.
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1420787
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So the height (x+8) has to be greater than or equal to 50% more than the width (x).

How do you express "50% more than" as a fraction or decimal?
Original post by Carthaginian
I would really really really help but I suck at maths and have no idea what you're going on about.
Here's your free reply.


I appreciate the thought anyways:biggrin:
Original post by offhegoes
So the height (x+8) has to be greater than or equal to 50% more than the width (x).

How do you express "50% more than" as a fraction or decimal?


half? So do i do this: (x+8)>1/2x
Original post by Chelsea12345
The diagram shows a rectangular birthday card which is x cm wide and (x + 8) cm tall. Given that the height of the card is to be at least 50% more than its width, a show that x ≤ 16.
Given also that the area of the front of the card is to be at least 180 cm2 , b find the set of possible values of x.
Could someone help me to answer this question?


a) Since the height is at least 50% of the width:

x + 8 >= x + x/2
x - x + 8 >= x/2
8 >= x/2
x <=16 proved

b) (x + 8) * x = 180
x^2 + 8x - 180 = 0
Using a GDC: x = 10 or x = -18

x = (-∞ , -18] U [10 , ∞)

but since x is cannot be a negative number because it is a side

x = [10 , ∞)
(edited 7 years ago)
Original post by Chelsea12345
half? So do i do this: (x+8)>1/2x


Hey I was thinking of making a whatsapp/kik/skype group a-level maths, would u care joining. I could lend a hand, and y I am predicted A at AS
Original post by ARK_REVISES
Hey I was thinking of making a whatsapp/kik/skype group a-level maths, would u care joining. I could lend a hand, and y I am predicted A at AS


yes I think so as long as i got time. i'm already piled up with overdue stuff at the moment.
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1420787
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Original post by Chelsea12345
half? So do i do this: (x+8)>1/2x


Nearly, but 50% more than x, so in terms of percentages that makes 150% of x.

As a decimal this is 1.5x.

x + 8 >= 1.5x
2x + 16 >= 3x <-- X2 both sides
16 >= x

For b), similar to the previous solution attempt posted. I'll update it.
Original post by Chelsea12345
yes I think so as long as i got time. i'm already piled up with overdue stuff at the moment.


so, it that an yes?
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Original post by dimapos32
a) Since the height is at least 50% of the width:

x + 8 >= x + x/2
x - x + 8 >= x/2
8 >= x/2
x <=16 proved

b) (x + 8) * x = 180
x^2 + 8x - 180 = 0
Using a GDC: x = 10 or x = -18


Nearly there, now draw a graph of x^2 + 8x - 180 = 0, with 10 and -18 as roots. We're looking for a range of values.
Original post by ARK_REVISES
so, it that an yes?


yep! are you in year 13 or year 12?
Original post by Chelsea12345
yep! are you in year 13 or year 12?


year 12,u?
Original post by ARK_REVISES
year 12,u?


same
Original post by Chelsea12345
same


so want use kik, whatsaap or something?
Original post by offhegoes
Nearly there, now draw a graph of x^2 + 8x - 180 = 0, with 10 and -18 as roots. We're looking for a range of values.


Oh my bad! I didnt see that the area was at least 180cm^2.

Hence,

x = (-∞ , -18] U [10 , ∞)

but since x is cannot be a negative number because it is a side

x = [10 , ∞)
(edited 7 years ago)

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