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
18 9,2 3,3,2 2 x 3 x 3 x c x c x c =18c3
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
2 x 3 x 3 x b x c = 18bc
2 x 3 x 3 x b x c = 18bc
-2
The other name of an anti-derivative is an integral. An integral is the function which finds the area under a line. Let me give you an example. The integral of xn would be (xn+1/n+1)+C. So the integral of x2 would be (x3/3)+C. If you wanted to know what the area under the equation x2 when x=3 (in other words, all values of x between 0 and 3 on the x-axis), then you can do the equation ((3)3/3)+C and area would be 3 units squared. C is the constant of integration. For more info on C, you can go to http://en.wikipedia.org/wiki/Constant_of_integration or simply look it up
limit x→3 1/c-1/3/c-3
The specific heat capacity of silver is 0.235 J/gĀ°C. To raise the temperature of 3 g of silver by 5 Ā°C (from 15 to 20Ā°C), you would need 3 g x 5 Ā°C x 0.235 J/gĀ°C = 3.525 J of energy.
C is 100, X is 10 and III is 3. CXIII is 213.
Integral of sqrt(2x) = integral of (2x)1/2 = √2/(3/2)*(x)3/2 + c = (2√2)/3*(x)3/2 + c where c is the constant of integration. Check: ( (2√2)/3*(2x)3/2 + c )' = (2√2)/3*(3/2)(x)(3/2)-1 + 0 = √2*(x)1/2 = √2x