Maths year 11

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    (Original post by K-Man_PhysCheM)
    For the first question, what if you use the upper bound of the shorter train (and the lower bound of the longer train)? Do you get an even smaller difference?

    The volume is the space that an object occupies. It's a bit like area, but for 3D objects, like cubes.

    To work out the area of a rectangle, you would multiply the lengths of its two perpendicular sides together. In a square, both side lengths are the same, so to find it's area you square the side length (multiply it by itself).

    Attachment 573814

    Volume brings this into 3-Dimensions: to calculate the volume of a cuboid, you need to multiply together the three side lengths; Volume (of cuboid) = a x b x c.

    In a cube, all sides are the same length, so you need to do Volume (of cube) = a x a x a = a^3 (side length cubed). The maximum volume will come from cubing the upper bound of the side lengths. These formulas will* be given to you at the front of the paper or on a separate formula sheet, so you don't need to memorise them: you can just check the sheet.

    EDIT: *Those formulas might be given to you, it will depend on your exam board though.
    So I can do

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    This is question 7



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    (Original post by z_o_e)
    So I can do

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    Ah, almost. Remember, to find the volume you are multiplying together the three side lengths: in the case of a cube, the three side lengths all have the same value (let's call it a), so you do a^3, which just means a x a x a. Your units would be cm^3. In this case, Volume = 40.5^3 = 40.5 x 40.5 x 40.5 .

    Or, you can use the "cube" button on your calculator. On my calculator, x^3 is written above the x^2 button, so I need to press "Shift" then "x^2" button to trigger the x^3 operation. It may be different on yours, so it would be best to ask a teacher at school. For now, just type in 40.5 x 40.5 x 40.5.
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    (Original post by K-Man_PhysCheM)
    Ah, almost. Remember, to find the volume you are multiplying together the three side lengths: in the case of a cube, the three side lengths all have the same value (let's call it a), so you do a^3, which just means a x a x a. Your units would be cm^3. In this case, Volume = 40.5^3 = 40.5 x 40.5 x 40.5 .

    Or, you can use the "cube" button on your calculator. On my calculator, x^3 is written above the x^2 button, so I need to press "Shift" then "x^2" button to trigger the x^3 operation. It may be different on yours, so it would be best to ask a teacher at school. For now, just type in 40.5 x 40.5 x 40.5.
    Yepp thank you x I've done cubes in school I got 66430.125



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    (Original post by z_o_e)
    This is question 7



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    Very good, question 7, part a, is correct! You might be asked to write the units, though. What might the units be, if you have found the volume from measurements in metres?

    Then try part b), which could be expressed in other words as: how many whole 5m^3 volumes fit inside the fish tank?
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    (Original post by z_o_e)
    Yepp thank you x I've done cubes in school I got 66430.125



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    Good. Just in case you are asked, what would the units be this time?
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    (Original post by K-Man_PhysCheM)
    Very good, question 7, part a, is correct! You might be asked to write the units, though. What might the units be, if you have found the volume from measurements in metres?

    Then try part b), which could be expressed in other words as: how many whole 5m^3 volumes fit inside the fish tank?
    Cm3

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    (Original post by K-Man_PhysCheM)
    Good. Just in case you are asked, what would the units be this time?
    Ummm wb this?

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    (Original post by z_o_e)
    Ummm wb this?

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    Yes, very good. Just be careful, the question asks for how many whole fish are allowed in the tank (or as I rephrased it, how many whole 5m^3 volumes fit in the tank), so you can't have a decimal number, because that's not a whole number.
    In this case, it might be tempting to round up, but DON'T!!. 1093 fish would be too many in the tank, and that many would not be allowed by the laws in the question. So the answer is 1092, as that is the maximum number of whole fish allowed in the tank.
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    (Original post by K-Man_PhysCheM)
    Yes, very good. Just be careful, the question asks for how many whole fish are allowed in the tank (or as I rephrased it, how many whole 5m^3 volumes fit in the tank), so you can't have a decimal number, because that's not a whole number.
    In this case, it might be tempting to round up, but DON'T!!. 1093 fish would be too many in the tank, and that many would not be allowed by the laws in the question. So the answer is 1092, as that is the maximum number of whole fish allowed in the tank.
    thank you SO MUCH!!!!

    Can you explain this please



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    (Original post by K-Man_PhysCheM)
    Yes, very good. Just be careful, the question asks for how many whole fish are allowed in the tank (or as I rephrased it, how many whole 5m^3 volumes fit in the tank), so you can't have a decimal number, because that's not a whole number.
    In this case, it might be tempting to round up, but DON'T!!. 1093 fish would be too many in the tank, and that many would not be allowed by the laws in the question. So the answer is 1092, as that is the maximum number of whole fish allowed in the tank.
    thank you SO MUCH!!!!

    Can you explain this please


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    (Original post by z_o_e)
    thank you SO MUCH!!!!

    Can you explain this please



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    OK, so when you see a question asking for the greatest or smallest something, you know you are probably going to need to use bounds. So the very first step is to find the upper and lower bounds of 5cm.

    The first part of the question asks for the greatest area of the circle. To find the greatest area, you need to use the biggest number, so you use the upper bound for the radius. Then you input the upper bound of the radius into the area of a circle equation: Area = radius^2 x pi (or A = r x r x pi). Use the pi button on your calculator, or 3.14159 if you don't have a pi button.
    Then, round the answer to three significant figures (take the first three digits, and then round off the 3rd digit accordingly, and get rid of any trailing zeroes that appear AFTER the decimal point).

    The second part asks for the smallest circumference. C = 2 x pi x radius. You want the smallest circumference, so use the lower bound of the radius. Then round your answer to 3 sig figs. Include units for both.

    Numerical solution (only check once you've had a go):
    Spoiler:
    Show
    UB = 5.5cm, LB = 4.5cm
    Max Area = 5.5^2 x pi = 95.0331... = 95.0cm^2 to 3 sf
    Min Circumference = 2 x 4.5 x pi = 28.27433... = 28.3cm to 3sf
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    (Original post by K-Man_PhysCheM)
    OK, so when you see a question asking for the greatest or smallest something, you know you are probably going to need to use bounds. So the very first step is to find the upper and lower bounds of 5cm.

    The first part of the question asks for the greatest area of the circle. To find the greatest area, you need to use the biggest number, so you use the upper bound for the radius. Then you input the upper bound of the radius into the area of a circle equation: Area = radius^2 x pi (or A = r x r x pi). Use the pi button on your calculator, or 3.14159 if you don't have a pi button.
    Then, round the answer to three significant figures (take the first three digits, and then round off the 3rd digit accordingly, and get rid of any trailing zeroes that appear AFTER the decimal point).

    The second part asks for the smallest circumference. C = 2 x pi x radius. You want the smallest circumference, so use the lower bound of the radius. Then round your answer to 3 sig figs. Include units for both.

    Numerical solution (only check once you've had a go):
    Spoiler:
    Show
    UB = 5.5cm, LB = 4.5cm
    Max Area = 5.5^2 x pi = 95.0331... = 95.0cm^2 to 3 sf
    Min Circumference = 2 x 4.5 x pi = 28.27433... = 28.3cm to 3sf


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    You haven't used the bounds.
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    Sorry just corrected it!

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    Sorry just corrected it!

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    (Original post by RDKGames)
    You haven't used the bounds.
    Just done it hahaha sorry

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    (Original post by RDKGames)
    You haven't used the bounds.
    So this one would be 1/7?



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    (Original post by z_o_e)
    So this one would be 1/7?



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    Very close. I assume you can see what happens with the denominator since you got that right, but what happens on the numerator?

    Also it wouldn't make sense for \frac{1}{\sqrt{7}} to be the same as \frac{1}{7}. When you rationalise something like that, the quantity remains the same but you are rationalising the denominator.
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    (Original post by RDKGames)
    Very close. I assume you can see what happens with the denominator since you got that right, but what happens on the numerator?

    Also it wouldn't make sense for \frac{1}{\sqrt{7}} to be the same as \frac{1}{7}. When you rationalise something like that, the quantity remains the same but you are rationalising the denominator.


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