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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
AXC and BXC form a linear pair.?
If angles AXC and BXC form a linear pair, it means they are adjacent angles that share a common vertex (point X) and their non-common sides (rays AX and BX) form a straight line. Consequently, the measures of angles AXC and BXC add up to 180 degrees. This property is fundamental in geometry, indicating that the two angles are supplementary.
How do you solve for x in ax-x equals c?
ax - x = c Factor 'x' Hence x(a - 1) = c (NB '1' is ONE , not 'I'. ) Divide both sides by 'a-1' Hence x = c/(a-1) Solved for 'x'.
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
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
AXC and BXC form a linear pair.?
If angles AXC and BXC form a linear pair, it means they are adjacent angles that share a common vertex (point X) and their non-common sides (rays AX and BX) form a straight line. Consequently, the measures of angles AXC and BXC add up to 180 degrees. This property is fundamental in geometry, indicating that the two angles are supplementary.
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
How do you solve for x in ax-x equals c?
ax - x = c Factor 'x' Hence x(a - 1) = c (NB '1' is ONE , not 'I'. ) Divide both sides by 'a-1' Hence x = c/(a-1) Solved for '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
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 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
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
In what year did BlueLinx Holdings Inc - BXC - have its IPO?
BlueLinx Holdings Inc. (BXC)had its IPO in 2004.
