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First assess that the initial case of where n = 1 is true. In this case, 6+9=15=53, so we can see that the case is true.We can then assume that 6k+9 is a multiple of 5, so we can let 6...
An important rule to remember when dealing with complex numbers is that if the complex number a+bi is a root of a polynomial, then the complex conjugate (a-bi) must also be a root of this polynomial.There...
32x(4x^2 + 3)^3
4sinx(cosx)^3 - 4cosx(sinx)^3
First we need to find the complementary function - the solutions of y'' -3y' + 2y = 0. By setting y=Ae^(kx) (A is a constant) we get :A(k^2 - 3k + 2)y = 0. This means k^2 - 3k + 2 = 0 =(k-1)(k-2). If 2 th...
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