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shubhro
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#1
A's age is equal to the sum of the ages of B and C. Ten years ago A was twice as old as B. Show that 10 years hence A will be twice as old as C.
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shubhro
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#2
Plantagenet Crown
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#3
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#3
Hang on, do you want us to answer these or John, because you're putting it in the Q and A section...?
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BlueSam3
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#4
shubhro
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#5
you may like to play around with – 7x2y2.
check out if it reveals a hidden pattern
check out if it reveals a hidden pattern
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shubhro
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#6
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brianeverit
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#7
shubhro
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#8
(x4 + y4)– 7x2y2
what would you need to make (x4 + y4) into a perfect square.
what would you need to make (x4 + y4) into a perfect square.
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BlueSam3
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#9
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#9
(Original post by brianeverit)
Surely this will NOT factorise.
Surely this will NOT factorise.
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contrapositive
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#10
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#10
(Original post by brianeverit)
Surely this will NOT factorise.
Surely this will NOT factorise.
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shubhro
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#11
contrapositive
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#12
shubhro
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#13
if you are still missing it by a whisker, may be you could take a look at the answer
http://ipm-mtse-olympiad-state-schol...torize_24.html
http://ipm-mtse-olympiad-state-schol...torize_24.html
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shubhro
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#14
a number when divided by 123 leaves a remainder of 89. what is the remainder when it is divided by 41?
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shubhro
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#15
how to solve this quadratic equation
Solve for x:
x2/3 + 3/ x2 = 5(x/3 – 1/x)
Solve for x:
x2/3 + 3/ x2 = 5(x/3 – 1/x)
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Hamushte
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#16
Matureb
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#17
Exon
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#18
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#18
If you expand the brackets and multiply through by 3x2, you will find yourself with a quartic not a quadratic.
That's all I have to contribute I'm afraid.
That's all I have to contribute I'm afraid.
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brianeverit
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#19
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#19
(Original post by BlueSam3)
It will. Surprisingly nicely, actually.
It will. Surprisingly nicely, actually.
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brianeverit
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#20
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#20
(Original post by Exon)
If you expand the brackets and multiply through by 3x2, you will find yourself with a quartic not a quadratic.
That's all I have to contribute I'm afraid.
If you expand the brackets and multiply through by 3x2, you will find yourself with a quartic not a quadratic.
That's all I have to contribute I'm afraid.
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