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Anonymous asked: Since you spell maths with an 's', are you British?

doers it matters?

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Anonymous asked: Is 4/9 greater than 1/3?

that is where lcm comes handy

lcm of 9 and 3 is 9

multiply it to both the numbers

4/9 *9 = 4

1/3*9 = 3

so 4/9 is greater

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can u give me the solution for -

the value of cube root of 64 raised to the power of -2

cube root of 64 is 4

4 raised to power to -2 = 1/16

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emeraldbeaches asked: thanks. b would be 2/3x or just 2/3 without the x?

2/3x

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please help, im clueless about this. I know this has something to do with binomial expansion and I expanded 3+nx^2 but that’s all I have done. also expand (1+2/3x)^n by the formula of binomial expansion multiply constant terms and equate it to 9 of r.h.s. multiply and get the terms containing x, and equate it to 84x you will get value of n

please help, im clueless about this. I know this has something to do with binomial expansion and I expanded 3+nx^2 but that’s all I have done.

also expand (1+2/3x)^n

by the formula of binomial expansion

multiply constant terms and equate it to 9 of r.h.s.

multiply and get the terms containing x, and equate it to 84x

you will get value of n

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52 notes &

vendittelli: Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?—- Paul R. Halmos [x] yeah

vendittelli:

Don’t just read it; fight it! Ask your own questions,
look for your own examples, discover your own proofs.
Is the hypothesis necessary? Is the converse true?
What happens in the classical special case? What
about the degenerate cases? Where does the proof
use the hypothesis?

—- Paul R. Halmos [x]

yeah

(via alainalovesmath)

6 notes &

darkeningofheart: askmemathsquestions: darkeningofheart: askmemathsquestions: Hi can you please explain this to me? I used the double angle formulae but I can’t seem to get past substituting 2cosx for sin2x 2cosx = 2sinxcosx or cosx-sinxcosx = 0 or cosx(1-sinx)=0 or cosx=0 or sinx =1 find the values of x in the interval given above, and you will get your answers. What happened to the 2’s before cosx and sinx? 2 got canceled out. Did they get cancelled out by dividing or subtracting? Can you separate them from cosx and sinx? yes they got divided and cancelled out its like 2* a = 2*b a being cos x and b being sinx cosx just for example what will you get? 2 cancels out, and you get, a=b

darkeningofheart:

askmemathsquestions:

darkeningofheart:

askmemathsquestions:

Hi can you please explain this to me? I used the double angle formulae but I can’t seem to get past substituting 2cosx for sin2x

2cosx = 2sinxcosx

or cosx-sinxcosx = 0

or cosx(1-sinx)=0

or

cosx=0 or sinx =1

find the values of x in the interval given above, and you will get your answers.

What happened to the 2’s before cosx and sinx?

2 got canceled out.

Did they get cancelled out by dividing or subtracting? Can you separate them from cosx and sinx?

yes

they got divided and cancelled out

its like 2* a = 2*b

a being cos x and b being sinx cosx just for example

what will you get?

2 cancels out, and you get, a=b

0 notes &

What is the least positive integer n such that

1 + 2 + … + n > 999

 

40

45

50

55

 

 

Thanch you!

 

 

sum to n natural numbers is

n(n+1)/2

given

n(n+1)/2 >999

solve this inequality

I think the answer is 44

6 notes &

darkeningofheart: askmemathsquestions: Hi can you please explain this to me? I used the double angle formulae but I can’t seem to get past substituting 2cosx for sin2x 2cosx = 2sinxcosx or cosx-sinxcosx = 0 or cosx(1-sinx)=0 or cosx=0 or sinx =1 find the values of x in the interval given above, and you will get your answers. What happened to the 2’s before cosx and sinx? 2 got canceled out.

darkeningofheart:

askmemathsquestions:

Hi can you please explain this to me? I used the double angle formulae but I can’t seem to get past substituting 2cosx for sin2x

2cosx = 2sinxcosx

or cosx-sinxcosx = 0

or cosx(1-sinx)=0

or

cosx=0 or sinx =1

find the values of x in the interval given above, and you will get your answers.

What happened to the 2’s before cosx and sinx?

2 got canceled out.

6 notes &

Hi can you please explain this to me? I used the double angle formulae but I can’t seem to get past substituting 2cosx for sin2x 2cosx = 2sinxcosx or cosx-sinxcosx = 0 or cosx(1-sinx)=0 or cosx=0 or sinx =1 find the values of x in the interval given above, and you will get your answers.

Hi can you please explain this to me? I used the double angle formulae but I can’t seem to get past substituting 2cosx for sin2x

2cosx = 2sinxcosx

or cosx-sinxcosx = 0

or cosx(1-sinx)=0

or

cosx=0 or sinx =1

find the values of x in the interval given above, and you will get your answers.

0 notes &

increasetheintensity asked: are you online right now? o:

not if you are taking a math test