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Type: Posts; User: mrenney
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How does (bsubj bsubk... bsubm)(bsubj bsubk... bsubm) become B at the end, wouldn't it still be B^2?
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where SL(R,2) is the matrix [a b] | a, b, c, d belongs to the set of reals, ad - because = 1.
[c d]
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Show that {Q( square root of 13,+,*} is a field?
Let Q( square root of 13 ={x+y square root of 13 where x,y are elements of Q}
Q of course is rational numbers
Addition and multiplication are...
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Do you think you can help me out
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Let f(a) = r(n) a^n +... + r(m) a^m and g(a) = r"(s) a^s +...+ r" (q) a^q be in R|a| with m>n, s>q and neither r(m) nor r"(q) equal to 0. Then f(a) g(a) = `r(n+s) a^(n+s) +...`r(m+q) a^(m+q). Notice...
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Show that {Q( square root of 13,+,*} is a field?
Let Q( square root of 13 ={x+y square root of 13 where x,y are elements of Q}
Q of course is rational numbers
Addition and multiplication are...
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I got kicked offline at my school's wifi, could you help with the problem now?
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Show that {Q(√13,+,*} is a field?
Let Q(√13 ={x+y√13 where x,y are elements of Q}
Q of course is rational numbers
Addition and multiplication are defined as follows
(a+b√13)+(c+d√13)...
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that if a^3 divides b^2, then a divides b. (Hint: Write a and b as their prime factorizations, a = p sub1 raised to the e sub1... p subn raised to the e subn and b = p sub1 raised to the d sub1... p...
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Why an even n gives you 2n petals and an odd m gives you m petals on polar grids?
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that has the following characteristics:
At t=0, the curve starts at the point (-4,4), travels to (3,-1) along an interesting non linear path, and returns to (-4,4) along a curved path that is a...
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and 2sin^2(x)cos(x) = cos(x)?
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and b. cos(sin^-1(x) - cos^-1(y))
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Sorry I wrote this question wrong again this is really the correct one, verify this identity: (sin^3(x) - cos^3(x)) / (sin(x) + cos(x)) = (csc^2(x) - cot(x) - 2cos^2(x)) / (1 - cot^2(x))
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Sorry I wrote the question wrong this is the correct one, verify this identity: (sin^3(x) cos^3(x)) /(sin(x) + cos(x)) = (csc^2(x) - cot(x) - 2cos^2(x)) / (1 - cot^2(x))
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= (csc^2(x) - cot(x) - 2cos^2(x))/1 - cot^2(x)
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