Why is anything to the 0th power 1? (answer)
No, here is one way you can think of it: Remember the rule that: xm/xn = xm-n (if x≠0) So if m = n, this says: xm/xm = x0 But clearly xm/xm = 1, so we must have that x0 = 1 as well. Be...
No, here is one way you can think of it: Remember the rule that: xm/xn = xm-n (if x≠0) So if m = n, this says: xm/xm = x0 But clearly xm/xm = 1, so we must have that x0 = 1 as well. Be...
As a quick check, if we are to have any hope of factoring a quadratic equation (of the form ax2+bx+c = 0) into linear factors with rational coeffecients we must have that the discriminant of the quadratic equation (which is defined to be: b2 - 4ac) is a perfect...
I'm not sure from what approach you are being asked to solve this question (e.g. linear algebra, precalculus, etc.), but I'll assume it's something along the lines of precalculus. Since we have only 1 equation and 4 unknowns, we simply solve for any 1 variable, let's say a: a...
This is actually a lot more intuitive than it seems. For 1-dimensional derivatives, we have that the best linear approximation of f(x) near x=a is: f(x) ≈ f(a) + f'(a) * (x-a) We have the same thing for higher dimensions: R(x) ≈ R(a) +...
I'm a little unsure as to what you are asking about. Are you asking why/how we get this differential equation? If that is your question, I'll try to help. If we use Kirchhoff's voltage law, we have that: sum of potential differences = 0. If we assign,...
Well...I don't know your background, and so I'm not sure whether you would like just a formula here or a derivation of a formula. I'll give a derivation using difference quotients (which you can look up if you'd like). First, some assumptions: You deposit money at the beginning of the...
1. I'm not sure what you meant by that first ^. I'll assume that it was the 9th root of 16, i.e. 16^(1/9) or 9√(16). In that case, we have (notice when the exponents multiply because we are raising numbers to powers and when they add because we are multiplying numbers with the same base): (...
Well...one may see that each term can be represented as the "cube" of a simpler term, i.e. we may write: u3 - 8w3 = (u)3 - (2w)3 Then, using the formula known as the "difference of cubes", which states: a3 - b3 = (a-b) * (a2 +...
Well...if you are referring to: ∂2f/(∂y∂x) then you just take the partial derivatives one at a time: First use the product rule: ∂f/∂x = ∂/∂x[ (xy) * (e^(x2y)) ] = ∂/∂x[xy]*(e^(x2y)) + ∂/∂x[e^(x2y)]*(xy) = ye^(x2y) + 2xye^(x2y)(xy) = e^(x2y) * (y...
So I'm a little unclear about the statement of the question, but I'll assume we are to show how transformations can be applied to f(x) = x2 to obtain g(x), h(x), i(x), and j(x). g(x) = (x-2)2 is f(x) = x2 shifted 2 units to the right (because we substitute x with x-2),...