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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
What is the factor of x2y-xyb-abx plus ab2?
(b - x)(ab - xy)
Can someone explain how to get x by itself in the equation x over a plus x over b equals 1 I know there answer is ab over a plus b but i don't know why?
x/a + x/b = 1Taking x as a common factor: x*(1/a + 1/b) = 1Adding the fractional parts x*(b/ab + a/ab) = 1or x*[(b + a)/ab] = 1Multiply both sides by (a+b)/abx = (a+b)/abAnother Approuch:-x/a+x/b = 1Multiply all terms by ab to eliminate the fractions:bx+ax = abFactorise:x(a+b) = abDivide both sides by (a+b):x = ab/(a+b)
Why do a square's diagonals bisect each other?
Consider the square ABCD whose diagonals, AC and BD, meet at X.AB is parallel to DC and AC intercepts them. Therefore <BAX = <DCX. AB is parallel to DC and BD intercepts them. Therefore <ABX = <CDX. AB = CD. Therefore triangle ABX is congruent to triangle CDX (SAS). So AX = XC ie X is the midpoint of AC and BX = XB ie X is the midpoint of BD. ie the diagonals bisect each other.
What is exponent growth?
That means that the growth is equal to, or similar to, an exponential function, which can be written (for example) as abx, for constants "a" and "b". One characteristic of exponential growth is that the function increases by the same percentage in the same time period. For example, it increases 5%, or equivalently by a factor of 1.05, every year.
