Find all the values of k for which the function y=sin(kt) satisfies the differential equation y’’ +14y=0?
Differentiate the proposed solution twice: y = sin(k*t) y' = k*cos(k*t) y'' = -(k^2)*sin(k*t) = -(k^2)*y Now plug this and the proposed solution into the differential equation: -(k^2)*sin(k*t) + 14*sin(k*t) = 0 By inspection, k = 0 is one solution, (the trivial solution of y = 0). Now divide through by sin(k*t): 14 - k^2 = 0 Now solve for k: k = +sqrt(14) and k = -sqrt(14) The differential equation is satisfied if k = 0, +sqrt(14), and -sqrt(14).
Determine the rate law and the value of k for the following reaction using the data provided...?
2N2O5(g)----> 4NO2(g) + O2(g) [N2O5]i (M) Initial Rate(M^-1 s^-1) 0.093 4.84x10^-4 ---- (1) 0.186 9.67x10^-4 ----- (2) 0.279 1.45x10^-3 ----- (3) From equation (1) & (2) it is evident that when [N2O5}i is doubled the initial rate is doubled, which implies the rate is directly proportional to [N2O5]. Similarly comparing equation (1) & (3) we observe that when [N2O5] is tripled the rate is also tripled. Hence the rate equation is Rate = k [N2O5] Using the data of any equation, say (1), we get 4.84x10^-4 = k x 0.093 OR k = 4.84x10^-4/0.093 = 5.2 x 10^-3 s-1 Hence the rate law is Rate = 5.2 x 10^-3 s-1[N2O5]
Math problem?? Find value of k?
k is an integer between 50 and 90 and is a multiple of 4. When k is divided by 5, the remainder is 3. When k is divided by 3, the remainder is 2. What is the value of k? The answer is 68. How did they get that?? Please explain.
How do I obtain constant terms in binomial expansion?
Let us try to view the general term of a binomial expansion in a slightly different way.Let us consider an example where we need to find the constant term in the expansion of [math](x - \frac{2}{x^2})^9[/math]General term for the above binomial is: [math]T_{r+1} = \ ^{9}C_{r}(x)^{9- r}(-\frac{2}{x^2})^{r}[/math][math]T_{r+1} = \ ^{9}C_{r}(x)^{9- r}(-2)^{r}(x)^{-2r}[/math][math]T_{r+1} = \ ^{9}C_{r}(-2)^{r}(x)^{9-3r}[/math]Now for a term in the expansion to be constant, the power of x should be 0.So, 9 - 3r = 0r =3Therefore, the 4th term in the expansion of [math](x - \frac{2}{x^2})^9[/math] is the constant term.[math]T_{3+1} = \ ^{9}C_{3}(-2)^{3}(x)^{9-3*3}[/math][math]T_{4} = \ 84 (-8) \ = \ - 672 [/math]Method: Step 1: Find the general term in the expansion of the binomial.Step 2: Collect all the powers of x terms and make it one entityStep 3: Set the power of x equal to 0 and find the value of rStep 4: Substitute back the value of r in the general term to get the constant term.Note: If r is fractional, then there is no constant term in the expansion.I hope it helps!
Finding complex roots of polynomial?
Solve 5z^4 - z^3 + 4z^2 - z +5 = 0, given the equation has no real roots. z^4 - 2z^3 + kz^2 - 18z + 45 = 0, has an imaginary root. Obtain all the roots and the value of k Please explain your method
How to write quadratic equation given a table of values?
This one can be done by inspection if you make a couple of assumptions. Since 3 points uniquely determine a parabola, you need to check that the extra 3 points given lie on the parabola found by inspection. Inspection: vertex form of parabola : y = a (x-h)^2 + k Assume that the data is consistent with a single quadratic function of the form y = a x^2 + k; where (h,k) is the vertex of the parabola. The data shows that (0,-100) is the vertex of the parabola: y = ax^2 - 100 Use (3,8) to find a: 8 = a(3)^2 - 100 : a = -12; Check that the other points lie satisfy y = -12x^2 - 100. [If they do not, then you need to use a least squares fit to find the value of a. but since this smells like a homework problem, a fit will probably be unnecessary.]
Calculus I Help - find the constant, k?
In order for the limit to exist at the crucial point x = 0, the value coming from the left (x < 0) must equal the value coming in from the right (x > 0). For x < 0, f(x) = sin(6x) which approaches f(0) = sin(6*0) = 0. So k must have the property that: [(2kx + sin(x)) / x] approaches 0 for x > 0. This next part if way beyond the scope of a course in Calc 1, but as x->0, (sin(x) / x) approaches 1. So (((2*k*x)) / x) must approach -1. Since the x terms cancel each other, 2k must equal -1 2k = -1 k = -1/2 . As I stated the fact when that the limit when x->0 implies that the limit of (sin(x)) / x approaches 0 is beyond a course in Calc 1. This is usually computed with l'Hopital's Rule (see Wikipedia) or power series which you won't see for a long time to come. .
How do I find the index of an XPath element in Selenium?
You can do it using count:count(parent::*/preceding-sibling::*) +1whereparent::* - Select the parent of the current node.preceding-sibling::* - Select the nodes which precedes the parent in its nodeset.Example : Here, i want to know the position of b node which holds value frd.xyzfrdOutput :count(a/b[.=’frd’]/preceding-sibling::*)+1
How do you get Python to print values rather than the memory addresses of functions?
If you pass the name of a function (or any other reference to a function) to the Python print statement (old Python) or the print() function (Python3 and Python2 after any from __future__ import print_function has been processed) … if you do this then you’ll see the object ID of that function (its address in the reference C implementation of the language).This is because that is the default string representation for all objects under Python.To get a value from any Python function (or method) you need to invoke the function by passing it an argument list (in parentheses). This is true even if your function takes no arguments. So for example a function such as:def foo(): return 42… would be invoked as foo(). You’d print the value returned by that function using print(foo()) (under Python3 or after importing the print_function from the __future__ module in older versions of Python).Understand this distinction, between reference and invocation, is one of the most fundamental skills which much be mastered by programmers in just about any programming language. It’s specially important in using Python, Ruby, Javascript, and any modern scripting or programming languages which support function references as “first class” objects (values which can be passed around like data).