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Here we have an equation involving absolute values. As a general rule |a| = +a and |a| = -a. We can apply that to our RHS. In the first case we get that 3x + 4 = 3x - 11, however after we subtract 3x from...

Here z is a complex number, and therefore, z = a + bi. Thus, our equation becomes (a + bi)^2 = i. Expanding the brackets we get a^2 + 2abi + (b^2)(i^2) = i. Since i is the squ...

when integrating you must add one to the power and divide by the new power. so y=(5x^(2 + 1))/3 + (x^(1+1))/2 + 2x + c now substitute 1 for x and 3 for y so 3= (5(1)^(2 + 1))/3 + ((1)^(1+1))/2 + 2(1) +...

The key to answering this question is to recognise that a common substitution of u=sin(x) wont work straight away so we must write the integral in a different form. Knowing that cos(2x)=1-2sin^2(x), the s...

1/4sin(2x) + x/2 + c