i think it is zero substitute a value and see for eg. 2i^+3j^+4k^
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∙ 12y agox^a / x^b = x^(a-b)andx^a * x^b = x^(a+b)
There are many methods. A simple way to test that X lies between A and B isIf (A-X)*(B-X) is negative then X lies between them.If (A-X)*(B-X) is positive then X does not lie between them.If (A-X)*(B-X) is zero then X = A or X = B.There are many methods. A simple way to test that X lies between A and B isIf (A-X)*(B-X) is negative then X lies between them.If (A-X)*(B-X) is positive then X does not lie between them.If (A-X)*(B-X) is zero then X = A or X = B.There are many methods. A simple way to test that X lies between A and B isIf (A-X)*(B-X) is negative then X lies between them.If (A-X)*(B-X) is positive then X does not lie between them.If (A-X)*(B-X) is zero then X = A or X = B.There are many methods. A simple way to test that X lies between A and B isIf (A-X)*(B-X) is negative then X lies between them.If (A-X)*(B-X) is positive then X does not lie between them.If (A-X)*(B-X) is zero then X = A or X = B.
2 x a x a x a x a x b x b x b = 2a4b3
Associative: (a + b) + c = a + (b + c) (a x b) x c = a x (b x c)
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)
(x - a) + (x - a) + (b) = 2 (x - a) + (b) = x - a + x - a + b = 2x - 2a + b
A condensed form of a * a * b * b * b is a2 * b3.a * a * b * b * b = a2 * b3
X^â x X^b= x^â+b x^a divided by x^b = x^a+b (x^a)^b=x^ab x^0=1 x^-a=1/x^a
X^â x X^b= x^â+b x^a divided by x^b = x^a+b (x^a)^b=x^ab x^0=1 x^-a=1/x^a
X^â x X^b= x^â+b x^a divided by x^b = x^a+b (x^a)^b=x^ab x^0=1 x^-a=1/x^a
a2b3
x^a / x^b = x^(a-b)andx^a * x^b = x^(a+b)
It must be x*(x+1). To see this, suppose that there existed a smaller common multiple formed by taking a*x and b*(x+1), where a =/= b since multiplying by the same number won't give you a common multiple. Then we have a*x < x*(x+1) => a < (x+1) b*(x+1) < x*(x+1) => b < x => a*b < x*(x+1). Also, a*x = b*(x+1) => x = b/(a-b) & (x+1) = a/(a-b). Therefore x*(x+1) = a*b/(a-b)^2 < x*(x+1)/(a-b)^2 => (a-b)^2 < 1 => (a-b) < 1. The problem here is that this requires that a=b, which cannot be. Therefore, x*(x+1) is the smallest common multiple of both x and (x+1)
[a, b] : a ≤ x ≤ b [a, b) : a ≤ x < b (a, b] : a < x ≤ b (a, b) : a < x < b
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
16B3 = 2 x 2 x 2 x 2 x B x B x B24B4 = 2 x 2 x 2 x 3 x B x B x B x BGreatest Common Factor = 2 x 2 x 2 x 2 x B x B x B = 8B3