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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?
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    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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    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?
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    (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
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    (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
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    (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 , ∞)
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    (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
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    (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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    (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.
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    (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.
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    (Original post by ARK_REVISES)
    so, it that an yes?
    yep! are you in year 13 or year 12?
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    (Original post by Chelsea12345)
    yep! are you in year 13 or year 12?
    year 12,u?
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    (Original post by ARK_REVISES)
    year 12,u?
    same
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    (Original post by Chelsea12345)
    same
    so want use kik, whatsaap or something?
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    (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 , ∞)
 
 
 
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