6 balls to an over in cricket
3c^(2) -17c - 6 Factors to ( 3c + 1)(c - 6 ) NB When you apply FOIL to the bracketed terms. F ; 3c X c = 3c^(2) O ; 3c X -6 = -18c I ; 1 X c = c L ; 1 x -6 = -6 Collecting 'like' terms 3c^(2) - 18c + c - 6 3c^(2) - 17c - 6 ( As before).
C. Every other letter is a vowel A E I O. The ones in between are Z B Y So z is to b as y is to C. So I think it's C
C
C^ (2) - D^(2) Factors to (C -D )(C + D) If we apply FOIL to these bracketed terms. (C -D )(C + D), then we have F ; C^(2) O = CD I = -DC L = -D^(2) 'Stringing out' C^(2) + CD - DC - D^(2) NB Remember CD= DC ; just like 2 x 3 = 6 & 3 x 2 =6 Hence C^(2) + CD - CD _ D^(2) Adding terms we have C^(2) - D^(2) NB THe (+)CD - CD = 0 This is the inverse function, done to show how C^(2) - D^(2) factors. NB Remember two squared terms with a negative(-) between WILL Factor. However, two squared with a positive(+) between them does NOT factor. As a n example, take the Pythagorean Eq'n. h^(2) = x^(2) + y^(2) This does NOT factor . However, h^(2) - y^(2) = x^(2) Does factors to (h - y)(h + y) = x^(2) Hope that helps!!!!! d
Exercise : Write a program to solve quadratic equations (ax^2 + bx + c), using a subroutinesub quadratic(a as double, b as double, c as double, indicator as long, root1 as double, root2 as double) Input parameters are a, b, c. Output parameters are :indicator (= 0 , two identical roots, = 1, two distinct real roots, = 2 , complex roots ) For real roots : they are stored in "root1" and "root2". For complex roots : root1 = real part, root2 = imaginary part. Ans : DEFLNG I-N DEFDBL A-H, O-Z L20: INPUT " Coefficients a,b,c (a*x^2+b*x+c) "; a, b, c CALL quadratic(a, b, c, iflag, r1, r2) IF iflag = 0 THEN PRINT "2 Identical roots "; r1 ELSEIF iflag = 1 THEN PRINT "2 distinct real roots "; r1; r2 ELSE PRINT "Complex root" PRINT "real part "; r1; "imaginary part "; r2 END IF GOTO L20 END SUB quadratic (a AS DOUBLE, b AS DOUBLE, c AS DOUBLE, indicator AS LONG, root1 AS DOUBLE, root2 AS DOUBLE) rem rem Input parameters: rem a*x*x + b*x + c rem rem Output parameters: rem indicator = 0 two identical roots, = 1 two distinct real roots, = 2 , complex roots. rem root1, root2 stores the real and imaginary part when complex roots occur. rem discrim = b * b - 4. * a * c IF discrim > 0 THEN dummy = SQR(discrim) indicator = 1 root1 = (-b + dummy) / (2. * a) root2 = (-b - dummy) / (2. * a) ELSEIF discrim = 0 THEN indicator = 0 root1 = -b / (2. * a) root2 = root1 ELSE indicator = 2 dummy = SQR(-discrim) root1 = -b / (2. * a) root2 = dummy / (2. * a) END IF END SUB
6 balls to an over in cricket
6 Balls to an Over in Cricket
3rd Rock from the Sun - 1996 B-D-O-C- 6-7 was released on: USA: 12 December 2000
6 b in a O
How many bowls to an over in cricket. There are 6 bowls in an over in cricket.
The Cleveland Show - 2009 B-M-O-C- 3-18 is rated/received certificates of: Netherlands:6
I assume that the "diameter of a triangle" means the diameter of the smallest circle containing it. Assume the triangle has sides a, b and c and angles A, B and C (where angle A is opposite side a etc.). Calculate angle A from the law of cosines. Then: a^2 = b^2 +c^2 - 2*b*c*cos(A) so: A = arccos((b^2 + c^2 - a^2)/(2*b*c)) -------------------(1) Then, using the usual construction for finding the center, O, of the circumscribed circle (i.e. a perpendicular bisectors of each side is constructed and all three meet at point O) we can draw the radius from O to A. This will divide angle A into two smaller angles (in some cases one angle will be larger than A and the other will be negative, but they will still add up to angle A). Using simple trigonometric formulas we can calculate angle A by adding the two together. The result then is: A = arccos(b/(2*r)) + arccos(c/(2*r)) -----------------------------(2) If we then equate the right hand sides of (1) and (2) we get an equation involving a, b, c and r. These can be solved to find r in terms of a,b and c. I used the easy way using Maple and got the result shown below. r = abs[sqr(-b^4+2*b^2*c^2+2*b^2*a^2-c^4+2*c^2*a^2-a^4)*a*c*b]/ [ b^4-2*b^2*c^2-2*b^2*a^2+c^4-2*c^2*a^2+a^4] The diameter is, of course, 2*r. This answer checks the cases for an equilateral triangle and for a 3,4,5 right triangle, each of which is easy to do by much simpler methods.
The O-C- - 2003 The Chrismukkah That Almost Wasn't 2-6 is rated/received certificates of: Australia:PG USA:TV-PG
photosynthesis- 6H O + 6O under the presence of sunlight and clorophyll=C H O 2 2 6 12 6
The White Shadow - 1978 B-M-O-C- 3-9 was released on: USA: 2 February 1981
S = Stimulas O = Organism B = Behaviour C = Consiquence
According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 2 words with the pattern C-O-B-O-. That is, eight letter words with 1st letter C and 3rd letter O and 5th letter B and 7th letter O. In alphabetical order, they are: cookbook crowboot