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Help with maths? ?

by Guest65661  |  earlier

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i) Find two quadratic equations which have roots of 3 and 4

ii) show algebraically why x^2 + y^2 =4 is not a function

iii) The volume of a rectangular prism is 500cm^3, if all the dimensions are halved, we halved the volume. Discuss giving examples.

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  1. i)

    y=x²-7x+12

    and

    y=2x²-14x+24

    ii)

    x²+y²=4

    x²=4-y²

    x=±√(4-y²)

    two y values for the same x value so it's not a function.

    iii)

    Volume=length x width x height

    V=lwh

    500 cm³=lwh

    (lwh)/2=500 cm³/2=250 cm³


  2. 1. 0=(x-3)(x-4) or 0=(2x-8)(2x-6)

    2. rearrange to get:

       y=sqrt(4-x^2)

      since square rooting give two answer (positive and negative) you will get               two answers for y but functions can only have one answer for y.

  3. x^2 --(3+4)x + 3*4 = 0 Or x^2 -- 7x + 12 = 0

    x^2 + y^2 = 4 is an equation of a circle, centre (0, 0), radius 2.

    No, by halving the dimensions, volume gets 1/8 of its original value.

  4. i)  Any polynomial of second degree can be written as:

    y = (x - a)(x - b) where a and  b are the roots.

    Therefore, you can write you quadratic equation like this:

    y = (x - 3)(x - 4) = x² - 3x - 4x + 12 = x² - 7x + 12

    To find a second one you can simply multiply y = x² - 7x + 12 by any number you feel like, lets say 2:

    y = 2x² - 14x + 24

    ii)  For a function to be algebraic it must pass the vertical line test (i.e. we can't find an x that has two y values).

    Therefore, let x = 0:

    x² + y² = 4

    (0)² + y² = 4

    y = ± 2

    Therefore, we've found an x, namely x = 0, that has two y values.  Therefore, x² + y² = 4 is NOT a function.

    iii)  

    Let L represent the length of the base.

    Let W represent the width of the base.

    Let H represent the height of the rectangular prism.

    Therefore, the volume is represented by:

    V = LWH

    Now, if we divide the dimensions by two we get that:

    V = (L/2)(W/2)(H/2) = (1/8)LWH

    This calculation shows that the statement given is false.  Looking at our new volume formula above we see that when the dimensions are halved we get 1/8 of the original volume.

    Hope this helps!
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