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Did the Russians win the Russian Revolution?
Given that during the Revolutions Russians were fighting Russians, you could say that Russians won. You could also say that Russians lost.
Did Russians descend from Ancient Rome?
No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.No, the Russians did not descend from the ancient Romans. Anthropologists say that modern Russians are of Slavic descent.
What is a group of Russians called?
Russians.
Tan 9 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
What is a antonym of pale?
Tan Tan
What is Russophobia?
Fear of Russians Fear of Russians
Have the Russians returned to Russia?
The Russians are in Russia.
What is tan20tan32 plus tan32tan38 plus tan38tan20?
This may not be the most efficient method but ... Let the three angle be A, B and C. Then note that A + B + C = 20+32+38 = 90 so that C = 90-A+B. Therefore, sin(C) = sin[(90-(A+B) = cos(A+B) and cos(C) = cos[(90-(A+B) = sin(A+B). So that tan(C) = sin(C)/cos(C) = cos(A+B) / sin(A+B) = cot(A+B) Now, tan(A+B) = [tan(A)+tan(B)] / [1- tan(A)*tan(B)] so cot(A+B) = [1- tan(A)*tan(B)] / [tan(A)+tan(B)] The given expressin is tan(A)*tan(B) + tan(B)*tan(C) + tan(C)*tan(A) = tan(A)*tan(B) + [tan(B) + tan(A)]*cot(A+B) substituting for cot(A+B) gives = tan(A)*tan(B) + [tan(B) + tan(A)]*[1- tan(A)*tan(B)]/[tan(A)+tan(B)] cancelling [tan(B) + tan(A)] and [tan(A) + tan(B)], which are equal, in the second expression. = tan(A)*tan(B) + [1- tan(A)*tan(B)] = 1
What percentage of ethnic Russians are Slavic?
Answer: All ethnic Russians are Slavic.
Why are Russians better than anybody?
Because they are Russians
What was the Russians MIR?
the Russians MIR is a controll in space
If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
