(x - a) + (x - a) + (b) = 2 (x - a) + (b) = x - a + x - a + b = 2x - 2a + b
[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)
x=b-a
4x^4 or 4x*4x*4x*4x A formula would be (x^a)(x^b)= x^a+b x^a) / (x^b)= x^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)
[a, b] : a ≤ x ≤ b [a, b) : a ≤ x < b (a, b] : a < x ≤ b (a, b) : a < x < 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)
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