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It's ax squared x bx x c = 0 So (ax x ax) x bx x c = 0 This is the quadratic formula, then you'll move onto the quadratic equation.
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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
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
Ax plus bx plus c equals 0?
x = -c/(a+b), provided a+b is not 0
Ax plus b equals bx plus c then x equals?
Ax + B = Bx + C Ax - Bx = (C - B) x (A - B) = (C - B) x = (C - B) / (A - B)
Solve the equation ax-b equals c for x?
ax - b = c ax = b + c x = (b + c)/a
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
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
Ax plus bx plus c equals 0?
x = -c/(a+b), provided a+b is not 0
If a x b equals 2 b x c equals 24 c x a equals 3 and a b c are all positive what is the value of a plus b plus c?
7.5
Ax plus b equals bx plus c then x equals?
Ax + B = Bx + C Ax - Bx = (C - B) x (A - B) = (C - B) x = (C - B) / (A - B)
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).
If x plus b equals c find x?
X= (b-c)/a
What is the simplified solution to BC plus 0 equals f?
f = B x C
What is a x squared plus b x plus c equals 0 in C programming?
ax2+bx+c = 0 is the general form of a quadratic equation which normally has two solutions
Prove if a union c equals b union c and a intersect c equals b intersect c then a equals b?
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
Solve the equation ax-b equals c for x?
ax - b = c ax = b + c x = (b + c)/a
How is the square root of a number found?
Let x be the parameter to be taken square root. a = 0 b = x loop: c = (a+b)/2 if c*c > x then b = c else a = c Repeat until accurate enough result is obtained in c or until c*c equals x.
