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Maths question help

Hi,

My first question is workout 2 x 2.2 x 10^12 x 1.5 x 10^12 / 2.2 x 10^12 -1.5 x 10^12 - Give your answer to 3 significant figures.

Got a bit stuck on this, managed to get 9.43x10^13



My second question is : x = √(p+q/pq) work out X P = 4 x 10^8 Q = 3 x 10^6
Give your answer to two significant figures.
Can't get anywhere near...

Final question is convert 7.352 to a fraction (the last two is recurring)

Thanks,

If possible can you show me how you got the answer?

Ben
Reply 1
Original post by Ben_K
Hi,

My first question is workout 2 x 2.2 x 10^12 x 1.5 x 10^12 / 2.2 x 10^12 -1.5 x 10^12 - Give your answer to 3 significant figures.

Got a bit stuck on this, managed to get 9.43x10^13

If the divider only 2.2x10^12 then you can simplify to 3x10^12-1.5x10^12


My second question is : x = √(p+q/pq) work out X P = 4 x 10^8 Q = 3 x 10^6
Give your answer to two significant figures.
Can't get anywhere near...

Final question is convert 7.352 to a fraction (the last two is recurring)

Thanks,

If possible can you show me how you got the answer?

Ben


P+Q =400 x 10^6 + 3 x 10^6 =403 x 10^6
PQ=12 x 10^14
X=V(403/12) x 10^3/10^7
Reply 2
With recurring decimal to fraction questions, you'll want to equate them to some variable (x), find a way to cancel out the recurring part then rearrange for x.

x=7.35˙2˙x = 7.3\dot{5}\dot{2}


Because you have two recurring digits, you'll want to move the decimal two places to the right. Moving it one place to the right entails multiplication by 10 therefore moving it two places means multiplying by 102 = 100.


100x=735.25˙2˙100x = 735.2\dot{5}\dot{2}

Do you see a way to cancel out the recurring digits?
(edited 10 years ago)
Reply 3
Original post by Exon
With recurring decimal to fraction questions, you'll want to equate them to some variable (x), find a way to cancel out the recurring part then rearrange for x.

x=7.35˙2˙x = 7.3\dot{5}\dot{2}


Because you have two recurring digits, you'll want to move the decimal two places to the right. Moving it one place to the right entails multiplication by 10 therefore moving it two places means multiplying by 102 = 100.


100x=735.25˙2˙100x = 735.2\dot{5}\dot{2}

Do you see a way to cancel out the recurring digits?


You minus 1 x however, I get confused. on the .352 do you take away .052 or .352?

Edit: OHH i think i get it now thanks
Reply 4
Original post by Ben_K
You minus 1 x however, I get confused. on the .352 do you take away .052 or .352?


You take off all of x (7.35˙2˙7.3\dot{5}\dot{2}) from 100x.

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