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How do you solve the equation ax bxc?
The expression you presented is not an equation. Do you mean ax2 + bx = c? Do you mean to solve it for x? I'm assuming that's the case, but you need to be more clear on your question. To solve for x then, the technique to use is called completing the square: ax2 + bx = c Multiply both sides by a: a2x2 + abx = ac Add the square of b/2 to both sides: a2x2 + abx + (b/2)2 = ac + (b/2)2 We now have a perfect square on the left, simplify: (ax + b/2)2 = ac + b2 / 4 (ax + b/2)2 = (4ac + b2) / 4 And now solve for x: ax + b/2 = ±[(4ac + b2) / 4]1/2 ax + b/2 = ± √(4ac + b2) / 2 ax = [-b ± √(4ac + b2)] / 2 x = [-b ± √(4ac + b2)] / 2a
How do you solve for x in ax-x equals c?
ax - x = c therefore x*(a - 1) = c Provide a ≠1, divide both sides by a - 1 to give x = c/(a - 1) If a = 1 then the value of x is indeterminate.
How do i solve rank of a determinant?
Matrix inverses and determinants, square and nonsingular, the equations AX = I and XA = I have the same solution, X. This solution is called the inverse of A.
If x is between a and c how are ax and ac related?
If a,b, and c are positive a < x < b means ax < cx If a,b , and c are negative a < x < b means ax > cx
What To solve an equation of the form ax you need to divide by the?
Co variable
Solve the equation ax-b equals c for x?
ax - b = c ax = b + c x = (b + c)/a
How do you solve the equation ax bxc?
The expression you presented is not an equation. Do you mean ax2 + bx = c? Do you mean to solve it for x? I'm assuming that's the case, but you need to be more clear on your question. To solve for x then, the technique to use is called completing the square: ax2 + bx = c Multiply both sides by a: a2x2 + abx = ac Add the square of b/2 to both sides: a2x2 + abx + (b/2)2 = ac + (b/2)2 We now have a perfect square on the left, simplify: (ax + b/2)2 = ac + b2 / 4 (ax + b/2)2 = (4ac + b2) / 4 And now solve for x: ax + b/2 = ±[(4ac + b2) / 4]1/2 ax + b/2 = ± √(4ac + b2) / 2 ax = [-b ± √(4ac + b2)] / 2 x = [-b ± √(4ac + b2)] / 2a
What are all of the multiplicative identities?
AxB=BxA (AxB)xC=Ax(BxC) Ax(B+C)=AxB+AxC Ax1=A Ax0=0
How do you solve for x in ax-x equals c?
ax - x = c therefore x*(a - 1) = c Provide a ≠1, divide both sides by a - 1 to give x = c/(a - 1) If a = 1 then the value of x is indeterminate.
How can we find the coordinates of x in the intersection of line and curve?
For example, the equation of a line: y = ax + b. the equation of a curve: y = cx2 + dx + e ax + b = cx2 + dx + e (solve for x)
What is the derivative of ax?
let f(x) = ax if a is a constant, then f'(x) = a if a is not constant, then f'(x) = ax' + a'x
Solve for x and y. ax plus by equals a-b bx-ay equals a plus b plus b?
Your two equations are: AX + BY = A - B BX - AY = A + B + B Because you have four variables (A, B, X, Y), you cannot solve for numerical values for X and Y. There are a total of four answers to this question, solving each equation for X and Y independently. First equation: X = (A - B - BY)/A Y= (A - B - AX)/B Second equation: X = (A +2B +AY)/B Y = (BX - A - 2B)/A
How too solve linear equations?
First rearrange the linear equation to the form ax + b = cThen subtract b from both sides: ax = c - b Divide both sides by a: x = (c - b)/a
Factorize x2 plus bx plus ax plus ab?
x2+bx+ax+ab = x2+ax+bx+ab = x(x+a)+b(x+a) = (x+a)(x+b)
How do you differentiate y equals x minus a over x?
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
How do i solve rank of a determinant?
Matrix inverses and determinants, square and nonsingular, the equations AX = I and XA = I have the same solution, X. This solution is called the inverse of A.
In what year did BlueLinx Holdings Inc - BXC - have its IPO?
BlueLinx Holdings Inc. (BXC)had its IPO in 2004.
