Mr Calculus: plz my generals i need ur help to dis o.@laplacian,rechiez,ben,fx,smurfy,ortarico.e.t.c
i dont know but d answers to d ques are 16.78' & 31.1'.
plz i want d workings not only d answers

Mr Calculus: plz my generals i need ur help to dis o.@laplacian,rechiez,ben,fx,smurfy,ortarico.e.t.c
i dont know but d answers to d ques are 16.78' & 31.1'.
plz i want d workings not only d answers

Mr Calculus: plz guys help me out wit dis......
if 3sin2@=5sin2¥....
find @ & ¥ if(tan@=2tan¥)

Laplacian:
3*2sin@cos@=5*2sin¥cos¥
and
sin@/cos@=2sin¥/cos¥....
Multiply&divide both eqns respectively;
3sin^2@=10sin^2¥
and
3cos^2@=2.5cos^2¥
add both equatns;
3=10sin^2¥+2.5cos^2¥
or
3=7.5sin^2¥+2.5
or
0.5=7.5sin^2¥
or
1=15sin^2¥
or
sin¥=sqr(1/15)
from there u can find ¥ and @

Humphrey77: if 3 raise to power 1998 is divided by 5 what is the remainder

Humphrey77: if 3 raise to power 1998 is divided by 5 what is the remainder

Humphrey77: if 3 raise to power 1998 is divided by 5 what is the remainder

Humphrey77: @ lapacian show it

Humphrey77: if 3 raise to power 1998 is divided by 5 what is the remainder

Laplacian:
4

Humphrey77: if ,m and n are open set than show that the intersection is also open
































@ laplacian

smurfy:

Yeah. sin¥ = sqrt(1/15) gives ¥ = 14.9 and @ = 28.1.

That's exactly what I got above.

Humphrey77: sin2x+sinx is equal to 1 we know that sin2x is equal to 2sinxcosx. clearly; we have 2sinxcosx +sinx equal to 1 . so let sinx equal to y; by subtitution into

sin2x + sinx equal to 1 we obtain the equation: 4y"4-3y"2-2y+1 equal to zero ; by solving for y we obtain y equal to 1; clearly recall that sinx is equal to y so we have x eq
ual to arcsin(1) we have x equal to 90 so x is equal to 90

Laplacian:
guy am stil an amatur ooo, (not yet a topologist)..
Let a<m<A and b<n<B denot d two open sets...
suppos, on d contrary, dat their intersection, p, is a closed set...
Then we can find x and y such dat
x<=p<=y, where a, b<x and y<A, B
now suppose, without loss of generality, dat A<B....then we can always find d numbers z such dat
y<z<A, since z<A<B, z should b in their intersection, but y<z a contradiction, showing dat z is not in their intersection...hence our supposition dat p is a closed set is untenable....and d result follows

jackpot: Hi Sir Laplacian

how can a set be less than or equal to a point x?

Laplacian:
solvin d differential eqn u cannot get y=1,
suppose y=1, indeed y'=0, y"=0,
substitute into ur differential eqn,
4y"4-3y"2-2y+1=0 to obtain

0-0-2+1=0
or
-1=0

hw did u solve ur differential eqn?

jackpot: Hi Sir Laplacian

how can a set be less than or equal to a point x?

jackpot: If you're not cleared as to how the answer above is gotten, note that for any natural number n,

3⁴ⁿ=(3⁴)ⁿ = 81ⁿ = 10M+1, for some integer M.

Similarly,
3⁴ⁿ⁺¹=3(3⁴ⁿ)=30M+3
3⁴ⁿ⁺²=3²(3⁴ⁿ)=90M+9
3⁴ⁿ⁺³=3³(3⁴ⁿ)=270M+27.

Observe that 3¹⁹⁹⁸ can be written as 3⁴ⁿ⁺², where n=499.

Now, for some integer M,

3¹⁹⁹⁸=3⁴ⁿ⁺²=90M+9=18(5)M+(5+4)=5(18M+1)+4.

From, the last equation, it is obvious why 3 raised to power 1998 leaves a remainder of 4 when divided by 5.

Gracias.

Laplacian:
guy am stil an amatur ooo, (not yet a topologist)..
Let a<m<A and b<n<B denot d two open sets...
suppos, on d contrary, dat their intersection, p, is a closed set...
Then we can find x and y such dat
x<=p<=y, where a, b<x and y<A, B
now suppose, without loss of generality, dat A<B.... we can always find d numbers z such dat
y<z<A, since z<A<B, z should b in their , but y<z a contradiction, showing dat z is not in their intersection...hence our supposition dat p is a set is untenable....and d result follows

Humphrey77: good work! BUT U NEED TO GO AND STUDY YR NUMBER THEORY VERY WELL

Humphrey77: sin2x+sinx is equal to 1 we know that sin2x is equal to 2sinxcosx. clearly; we have 2sinxcosx +sinx equal to 1 . so let sinx equal to y; by subtitution into

sin2x + sinx equal to 1 we obtain the equation: 4y⁴-3y²-2y+1 equal to zero ; by solving for y we obtain y equal to 1; clearly recall that sinx is equal to y so we have x equal to arcsin(1) we have x equal to 90 so x is equal to 90

jackpot: ^you are assuming that all your sets are subsets of IR, the set of real numbers and you are also assuming that elements can be compared by your use of less than sign. ( < )

my question is, what if the sets are open in R²? Or what if they are domains in the Complex Plane?

Can your proof work for an arbitrary open set?

jackpot: This guy, e be like say u snuff a little alcohol, bah? Why are u quoting people and replying nonsense? I drafted a solution that will convince even non-mathematicians on how the answer is gotten. I am not a stereotyped mathematician.

A mere 5 seconds look at the question, I pictured the answer. Integral powers of 3 ends with the cycle of digits {3, 9, 7, 1}.



1998Mod4=2

and 3 raised to power of the residue class of 2 yields 9.

and 9Mod5=4.

Maybe thats how you want the solution look like?

SMH.