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Math Questions

 


adri
Well like Metalfreek, I'm starting a topic with some "math questions". Not all of them require a lot of work, some of them well you just have to see those to solve them...

Question 1:
Well because I had to use a table for this one, I created a webpage to display the table properly. Smile
--> Link

Question is simple, what's the number that should replace the Questionmark.
Skills needed: + - * /

Question 2:
http://www.frihost.com/forums/vt-108646.html#905508

Question 3:
http://www.frihost.com/forums/vt-108646.html#905707


PS: Also check this topic: http://www.frihost.com/forums/vt-106519.html, some nice and tricky questions there too...


Adri


Last edited by adri on Fri Jul 24, 2009 11:29 am; edited 3 times in total
metalfreek
Is 5 the answer?
fx-trading-education
answer is 7 because (8 + 13) / 3 = 7
adri
fx-trading-education got it right, you had to add up the first to numbers and divide them by the third number and then you get the fourth number.
Code:

8+6=14/2=7
9+6=15/3=5
16+18= 34/17=2
10+10=20/5=4
8+13=21/3=7


New Question will be added soon... Smile

Adri
adri
Question 2:

They asked the baker, Mr. Wis, how many breads he had sold on monday. The baker answered:

My first customer bought half of the stock + half a bread. My Second customer, bought half of what was left + half a bread. The third customer asked for half of the stock that was still available + half a bread to feed the ducks. The last Customer, Mr Salomon, wanted to buy a bread, but there were no more breads.

Mr Bread didn't sold any half breads. Can you tell me how many breads he sold?




Adri
Hogwarts
Pretty shonky baker if he only bakes 7 loaves o_o
adri
Question 3:

Prove that:

  • sin(a+b)*sin(a-b)=cos˛b-cos˛a
  • cos3a = 4cosła-3cosa


*Those two have nothing to do with eachother, they are two seperate things. Smile


Adri
_AVG_
The first one:

TO PROVE: sin(a+b)*sin(a-b)=cos^2b-cos^2a

sin(a+b)*sin(a-b)=(sinacosb+sinbcosa)(sinacosb-sinbcosa)
=sin^2acos^2b-sinasinbcosacosb+sinasinbcosacosb-sin^2bcos^2a
=cos^2b(1-cos^2a)-cos^2a(1-cos^2b)=cos^2b-cos^2acos^2b-cos^2a+cos^2acos^2b
=cos^2b-cos^2a

And the second one:

TO PROVE: cos3a = 4cos^3a-3cosa

cosa+isina=e^ia
cos3a+isin3a=e^3ia=(e^ia)^3=(cosa+isina)^3=cos^3a+3icos^2asina-3cosasin^2a-isin^3a
=cos^3a-3cosasin^2a+i(3cos^2asina-sin^3a)
Equating Real parts, cos3a=cos^3a-3cosasin^2a=cos^3a-3cosa(1-cos^2a)=cos^3a-3cosa+3cos^a
=4cos^3a-3cosa


Last edited by _AVG_ on Sat Jul 25, 2009 9:42 am; edited 1 time in total
metalfreek
adri wrote:
Question 3:

Prove that:

  • sin(a+b)*sin(a-b)=cos˛b-cos˛a
  • cos3a = 4cosła-3cosa


*Those two have nothing to do with eachother, they are two seperate things. Smile


Adri

#1
sin(a+b)*sin(a-b)
=(sina cosb + cosa sinb)*(sina cosb - cosa sinb)
=sin^2a * cos^2b - cos^2a * sin^2b
=(1 - cos^2a) * cos^2b - cos^2a * (1-cos^2b)
=cos^2b - cos^2a*cos^2b - cos^2a + cos^2a*cos^2b
=cos˛b-cos˛a
proved#

#2 cos3a=4cosła-3cosa
cos3a
=cos(a+2a)
=cosacos2a-sinasin2a
=cosa(2cos^2a-1)-sina(2sinacosa)
=2cos^3a-cosa-2sin^2acosa
=2cos^3a-cosa-2cosa(1-cos^2a)
=2cos^3a-cosa-2cosa+2cos^3a
=4cosła-3cosa
proved#

Very Happy
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