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    I'm reallllyy bad at math and I've attempted to do these but it just goes over my head, I've never been able to do math. Anyways, could anyone work out the following with steps so I can understand how to do it? Thanks!

    A stone is thrown vertically upwards with a velocity of 20m s-1 is at a height y metres where y - 20t - 5t^2

    For how long is it over 10m above the ground?

    A batsman hits a cricket ball so that its trajectory has the equation y = x-1/80* x^2
    where x and y are horizontal and vertical distances measured in metres. How far away does it land and what is its greatest height?

    The Time t, an experiment takes depends on the temperature T degrees C and is given by the formula t=t^2-t+1
    Calculate the time the experiment takes when the temperature is a) -2C b) 5C

    If the time taken is 2 seconds find the temperature using an appropriate method. State T correct to 2 decimal places.
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    (Original post by mrpopodopalaus)
    I'm reallllyy bad at math and I've attempted to do these but it just goes over my head, I've never been able to do math. Anyways, could anyone work out the following with steps so I can understand how to do it? Thanks!

    A stone is thrown vertically upwards with a velocity of 20m s-1 is at a height y metres where y - 20t - 5t^2

    For how long is it over 10m above the ground?
    You want to find y > 10 \Rightarrow 20t - 5t^2 > 10 \Rightarrow -10 + 20t - 5t^2 > 0, sketch the quadratic -10 + 20t - 5t^2, find the roots, between which two values of t is it greater than zero? The difference of those two t values if your length of time.
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    (Original post by Zacken)
    You want to find y > 10 \Rightarrow 20t - 5t^2 > 10 \Rightarrow -10 + 20t - 5t^2 > 0, sketch the quadratic -10 + 20t - 5t^2, find the roots, between which two values of t is it greater than zero? The difference of those two t values if your length of time.
    Oh ok thanks, wouldn't have thought of it that way.
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    (Original post by mrpopodopalaus)
    A batsman hits a cricket ball so that its trajectory has the equation y = x-1/80* x^2
    where x and y are horizontal and vertical distances measured in metres. How far away does it land and what is its greatest height?.
    Plot the graph of y = x-1/80x^2 - it's going to be a parabola/quadratic. It will start from the origin and the second root will be the distance it lands (imagine the graph being quite literally the path of the projectile), the maximum height will occur where? You tell me, what do you know about maximums/minimums of quadratic equations?
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    (Original post by Zacken)
    Plot the graph of y = x-1/80x^2 - it's going to be a parabola/quadratic. It will start from the origin and the second root will be the distance it lands (imagine the graph being quite literally the path of the projectile), the maximum height will occur where? You tell me, what do you know about maximums/minimums of quadratic equations?
    I know nothing, this is something we were "supposed" to be taught in GCSE, however with the fact that I was always in the lower set, they never bothered to teach us what we needed to know. Now I'm doing BTEC I'm struggling with the maths aspect of science because I never learnt what I need to know. Considering this is BTEC, minimal help is given.
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    (Original post by mrpopodopalaus)
    I know nothing, this is something we were "supposed" to be taught in GCSE, however with the fact that I was always in the lower set, they never bothered to teach us what we needed to know. Now I'm doing BTEC I'm struggling with the maths aspect of science because I never learnt what I need to know. Considering this is BTEC, minimal help is given.
    A quadratic of the form ax^2 + bx + c has a minimum/maximum point when x = -\frac{b}{2a} and the value of that function at that point is given by just plugging x = -\frac{b}{2a} into your quadratic.
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    (Original post by Zacken)
    A quadratic of the form ax^2 + bx + c has a minimum/maximum point when x = -\frac{b}{2a} and the value of that function at that point is given by just plugging x = -\frac{b}{2a} into your quadratic.
    ok o.O and what does the last equation that you put mean?
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    (Original post by mrpopodopalaus)
    ok o.O and what does the last equation that you put mean?
    So you have some random quadratic like 1 + 2x + x^2, the minimum/maximum value will occurs when x = -\frac{2}{2 \times 1} = -1, so the minimum/maximum value of the quadratic is 1 + 2(-1) + (-1)^2 (i.e: you plug the value of x that gives the minimum into the quadratic to get the minimum value of the quadratic).
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    (Original post by Zacken)
    So you have some random quadratic like 1 + 2x + x^2, the minimum/maximum value will occurs when x = -\frac{2}{2 \times 1} = -1, so the minimum/maximum value of the quadratic is 1 + 2(-1) + (-1)^2 (i.e: you plug the value of x that gives the minimum into the quadratic to get the minimum value of the quadratic).
    Oh I see
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    (Original post by mrpopodopalaus)
    Oh I see
    So how you you apply that to y = x - \frac{x^2}{80}?
 
 
 
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