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Y b x c w

Updated: 4/28/2022
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13y ago

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The next term is ' d ' .

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13y ago
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Related questions

What comes next y b x c w?

d


What is the next letter y b x c w?

' d ' is.


What is next letter Y B X C W?

D


If x equals a-b y equals b-c z equals c-a then the value of x plus y plus z x plus y plus z is?

x + y + z = 0 x = a - b, y = b - c, z = c - a, therefore a - b + b - c + c - a = ? a - a + b - b + c - c = 0


What would be the next letter in the line y b x c w?

The next term is ' d ' .


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).


What letter is symmetrical?

B,C,c,D,E,H,I,K,l,M,N,O,T,U,V,v,W,w,X,x,Y


Which letter comes next Y B X C W?

The next term is ' D '.


Why can't 4 two - digit addends be greater then 400?

404


Once you've completed the square how do you solve the equation?

It depends upon the equation that you are trying to solve. for example if you had x² + y² - 4x - 6y = 3 completing the square (in both x and y) would give you (x - 2)² - 4 + (y - 3)² - 9 = 3 at which point you could collect the numbers together on the right to give you (x - 2)² + (y - 3)² = 16 = 4² and then "read off" the centre of the circle at (2, 3) and its radius of 4. However, I suspect that you are asking regarding a quadratic, in which case you have got to something like (x - w)² - r = 0 Just proceed as normal with an equation to solve: (x - w)² - r = 0 → (x - w)² = r → x - w = ±√r → x = w ±√r ------------------------------------------------------------------ For a quadratic of the form ax² + bx +c this gives: ax² + bx + c = 0 → x² +(b/a)x + (c/a) = 0 completing the square: → (x + (b/2a))² - (b/2a)² + c/a = 0 → (x + (b/2a))² = b²/4a² - 4ac/4a² → (x + (b/2a))² = (b² - 4ac)/4a² → x + (b/2a) = ±√((b² - 4ac)/4a²) → x = -(b/2a) ±(√(b² - 4ac))/2a → x = (-b ±√(b² - 4ac))/2a which you may recognise as the formula for solving a quadratic equation.


What is pi to 12 decimal places?

3.141592653589 Formula: y = a a=½(a+b) b=√by c = c - x(a-y)2 x = 2x π = (a+b)24c


How can I solve a quadratic equations for x in terms of y?

A quadratic involving x and y is usually in the form 'y = ax2 + bx + c'. This form is y in terms of x, so we must rearrange it. y = ax2 + bx + c y/a = x2 + bx/a + c/a y/a = x2 + bx/a + d + e, where c/a = d + e, e = (b/a)2 y/a - e = x2 + bx/a + d y/a -e = (x + b/a)2 √(y/a - e) = x + b/a √(y/a - e) - b/a = x