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Six Balls to an Over in Cricket.
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∙ 15y ago
Wiki User
∙ 14y ago6 balls to an over in cricket
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What is the argument that A equals B B equals C so A equals C?
A=B , A-B=B-B , A-B =0 B=C , B-B=C-B, 0=C-B So A-B=0 but also C-B=0 A-B=C-B ...add +b ...A-B+B=C-B+B , A=C
What is the answer to A plus B plus C equals A times B equals C?
A=0 b=0 c=0
If you have a 10 measuring cup 4 measuring cup and a 3 measuring cup how many pours does it take to get the 4 to 4 the 3 to 1 and the 10 to half?
Assuming the 10 = Cup A, 4 = Cup B and 3 = Cup C 1) Fill Cup C (A=0, B=0, C=3) 2) Pour Cup C into Cup A (A=3, B=0, C=0) 3) Fill Cup B (A=3, B=4, C=0) 4) Fill Cup C from Cup A (A=3, B=1, C=3) 5) Pour the remainder of Cup B into Cup A (A=4, B=0, C=3) 6) Empty Cup C (A=4, B=0, C=0) 7) Fill Cup B (A=4, B=4, C=0) 8) Fill Cup C from Cup A (A=4, B=1, C=3) 9) Pour the remainder of Cup B into Cup A (A=5, B=0, C=3) 10) Empty Cup C (A=5, B=0, C=0) 11) Fill Cup B (A=5, B=4, C=0) 12) Fill Cup C from Cup A (A=5, B=1, C=3) 13) Empty Cup C (A=5, B=1, C=0) 13) Pour the remainder of Cup B into Cup C (A=5, B=0, C=1) 14) Fill Cup B (A=5, B=4, C=1) so assuming you count the filling of cups as pours your answer is 14
Does NOR gate obey associative law?
Presumably you mean is it true that: ( A nor B ) nor C == A nor ( B nor C ) ? No. Let's make a table: A B C (A nor B) (B nor C) [ (A nor B ) nor C ] [ A nor ( B nor C ) ] 0 0 0 1 1 0 0 0 0 1 1 0 0 1 .... So you see right away for A=0, B=0, and C=1 it doesn't work.
What 3 numbers give the same result when added or multiplied together?
The simplest answer is 1 and 2 and 3. 1+2+3=6 1x2x3=6 Other valid answers: 0, 0, 0 -1, 0, 1 -x, 0, x 1, 1.5, 5 The general formula here is: abc=a+b+c abc-a=b+c a(bc-1)=b+c a=(b+c)/(bc-1) If you choose two random numbers b and c you will ALWAYS have a number a that satisfies the conditions (unless bc=1)
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).
How many combinations of 6 numbers are there in 16 numbers What are these combinations?
There are 16C6 combinations = 16!/(6!(16-6)!) = 16!/(6!10!) = 8,008 possible combinations. Assuming the 16 numbers are the hexadecimal digits 0-F, they are: {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 6}, {0, 1, 2, 3, 4, 7}, ...., {0, 1, 2, 3, 4, E}, {0, 1, 2, 3, 4, F}, {0, 1, 2, 3, 5, 6}, ..., {0, 1, 2, 3, 5, F}, {0, 1, 2, 3, 6, 7}, ...., {0, 1, 2, 3, 6, F}, ...., {0, B, C, D, E, F}, {1, 2, 3, 4, 5}, ..., {9, A, B, C, D, E}, {9, A, B, C, D, F}, {9, A, B, C, E, F}, {9, A, B, D, E, F}, {9, A, C, D, E, F}, {9, B, C, D, E, F}, {A, B, C, D, E, F} I'll let you fill in the missing 7,990 possible combinations.
Determine the highest of the three input numbers using flowchart?
(start) /a=0 c=0\ \b=0 / /input a/ /input b/ /input c/ /a>b\ no /b>c\ yes /display b/ -> (a) \ / \ / yes no /a>c\ no /display c/ -> (a) \ / yes /display a/ <- (a) (end)
What is the negation of B is not between A and C?
The negation of B is not between A and C is = [(A < B < C) OR (C < B < A)] If A, B and C are numbers, then the above can be simplified to (B - A)*(C - B) > 0
A plus b equals c times c equals c times a?
well the only answer i can so far come up with is: a=1 b= -1 c=0 so then (a+b)=(c*c)=(a*c) ...............(0)=(0)=(0) ......................0 ok if that is what your looking for Great! just know that i am not a math wiz. :)
When xor and or operation are same exmple A or B or C equals A xor B xor C?
Check the following table: a b c a+b+c a^b^c 0 0 0 0 0 = 0 0 1 1 1 = 0 1 0 1 1 = 0 1 1 1 0 1 0 0 1 1 = 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 = So they are equal if the number of ones between a, b, and c is zero or an odd number.
If a equals positive and b equals negative and c equals zero how many solutions does a Quadratic Equation have?
Two: one is 0, the other is -b/a ax2 + bx + c = 0, but c = 0 ⇒ ax2 + bx + 0 = 0 ⇒ ax2 + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or (ax + b) = 0 ⇒ x = -b/a
