I don't quite understand the phrasing of the question. In mathematics a product is the result of multiplication. If you have three numbers A, B and C Then A x B might be one product A x C might be the second product B x C might be the third product This would leave A x B x C as the fourth product. So, if we have 3, 5 and 7 then 3 x 5 = 15 3 x 7 = 21 5 x 7 = 35 and 3 x 5 x 7 = 105
x^2 x^3? That would be x^5 If you mean x X 2x X 3, then what would be 2x^2+3
The general form is (2/3)*x^(3/2) + C
No. But multiplication is distributive over addition. This means that for any numbers A, B, and C A x (B + C) = (A x B) + (A x C). If addition were distributive over multiplication, that would mean that A + (B x C) = (A + B) x (A + C) which is not true.
18 x 24 = (18 x 20) + (18 x 4)
As a product of its prime factors: 3*3*3*3 = 81 and 4*3 = 12
Int = 3x^(2) dy y = 3x^(3) / 3 + c y = x^(3) + C
18 9,2 3,3,2 2 x 3 x 3 x c x c x c =18c3
sqrt(1) + 3*sqrt(x) = 1 + 3*x^1/2So the antiderivative is x + [3*x^(3/2)]/(3/2) + c = x + 2*x^(3/2) + c where c is the constant of integration.
I don't quite understand the phrasing of the question. In mathematics a product is the result of multiplication. If you have three numbers A, B and C Then A x B might be one product A x C might be the second product B x C might be the third product This would leave A x B x C as the fourth product. So, if we have 3, 5 and 7 then 3 x 5 = 15 3 x 7 = 21 5 x 7 = 35 and 3 x 5 x 7 = 105
3 x 7 x 37 = 777, a = 3 & c = 7
A Maclaurin series is centered about zero, while a Taylor series is centered about any point c. M(x) = [f(0)/0!] + [f'(0)/1!]x +[f''(0)/2!](x^2) + [f'''(0)/3!](x^3) + . . . for f(x). T(x) = [f(c)/0!] + [f'(c)/1!](x-c) +[f''(c)/2!]((x-c)^2) + [f'''(c)/3!]((x-c)^3) + . . . for f(x).
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Suppose y = c*x where c is the constant of proportionality. x = 3 and y = 30 so 30 = c*3 which implies that c = 10. So the equation is y = 10*x
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