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Mathematics: Post your doubts here!

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q7 c i ) use the same formula of volume of cone but use (3r)^2 instead of r^2 and use (3h) instead of h
it will be --> 1/3 x pie x (3r)^2 x 3h ---->1/3 x pie x 9r^2 x 3h -----> notice that --> 1/3 x pie x r^2 x h x 27 is equal to 1/3 x pie x 9r^2 x 3h SO equal 27 W

c ii ) use the same way as i
use the same formula of volume then instead of r^2 put (2x)^2 and instead of h put y
it will be --> 1/3 x pie x (2x)^2 x y ---> 1/3 x pie x 4x^2 x y ----> 4/3 x pie x x^2 x y ---->notice that --> 4(1/3) x pie x x^2 x y equals to 1/3 x pie x 4x^2 x y SO equals 4W
 
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q7 c i ) use the same formula of volume of cone but use (3r)^2 instead of r^2 and use (3h) instead of h
it will be --> 1/3 x pie x (3r)^2 x 3h ---->1/3 x pie x 9r^2 x 3h -----> notice that --> 1/3 x pie x r^2 x h x 27 is equal to 1/3 x pie x 9r^2 x 3h SO equal 27 W

c ii ) use the same way as i
use the same formula of volume then instead of r^2 put (2x)^2 and instead of h put y
it will be --> 1/3 x pie x (2x)^2 x y ---> 1/3 x pie x 4x^2 x y ----> 4/3 x pie x x^2 x y ---->notice that --> 4(1/3) x pie x x^2 x y equals to 1/3 x pie x 4x^2 x y SO equals 4W
what i didnt get is how did we consider 1/3 x pie x r^2 x h x 27 is equal to 1/3 x pie x 9r^2 x 3h SO equal 27 W???
 
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what i didnt get is how did we consider 1/3 x pie x r^2 x h x 27 is equal to 1/3 x pie x 9r^2 x 3h SO equal 27 W???
take 9 and 3 out of the equation to bring it back to its previous form then multiply all by 27 which is 9 by 3
you can check it if you make substitution with values of r and h in both equation (( i mean 1/3 x pie x 9r^2 x 3h and 1/3 x pie x r^2 x h x 27 )) and it will give you the same answer
 
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take 9 and 3 out of the equation to bring it back to its previous form then multiply all by 27 which is 9 by 3
you can check it if you make substitution with values of r and h in both equation (( i mean 1/3 x pie x 9r^2 x 3h and 1/3 x pie x r^2 x h x 27 )) and it will give you the same answer
THANK U SOOO MUCH
 

NIM

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Learn these 2 formulas:

1) This formula can only be applied if the difference is constant between the first and the second, second and third etc.
nth term = a + d (n - 1)
where:
'a' is the FIRST term
'd' is the DIFFERENCE between the terms


2) This formula can only be applied if, in simple terms, the first term is multiplied by a number to get the second term... and the second term is multiplied by the same number to get the third and so on:
nth term = ar^(n-1)
'a' is the FIRST term
'r' is the number they are being multiplied with
1, 2, 4, 8, 16 (for instance)
1*2^(n-1)
= 2^(n-1) Answer


There are many questions you can't solve even when you know these sequences. For that, you need to use some common sense and the sequences already given ;)
dude how abut "adding numbers with sequence"....
what should b the formula fr that one........:p
 
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its quite simple:
1) make 3 eqns. (since there are 3 variables)
2) they are:
a) -33+x=y
b) x+y=z
c) y+z=18
3) then replace y with -33 + x
i.e. x + -33 + x= z
and -33 + x + z =18
4) reorder and make it into a simultaneous eqn.
i.e. 2x-z=33 and x +z = 51
5) solve the eqn.
6) u will get the value of x, = 28
7) then put the value of x (28) in the first eqn. (a)
8) u will get y as -5
9) then put the value of y (-5) in the last eqn. (c)
10) u will get z as 23

so there u have it :D
hope it helps and best of luck!!!
 
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