With difficulty if you don't know the methods and one of the methods is as follows:- Let the points (k, 3h) and (3k, h) be (x1, y1) and (x2, y2) respectively: (y-y1) over (y2-y1) = (x-x1) over (x2-x1) (y-3h) over (h-3h) = (x-k) over (3k-k) (y-3h) over (-2h)) = (x-k) over (2k) Multiply both sides by -2hk to eliminate the fractions: k(y-3h) = -h(x-k) ky-3hk = -hx+hk ky = -hx+hk+3hk ky = -hx+4hk The above straight line equation can be expressed as: hx+ky-4hk = 0
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graph G(x)=[x]-1
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Points: (k, 3h) and (3k, h) Slope: h-3h/3k-k = -2h/2k => -h/k Equation: y-3h = -h/k(x-k) => ky-3hk = -h(x-k) => ky = -hx+hk+3hk Equation in its general form: hx+ky-4hk = 0
HX Draw was created in 1911.
HX Civics come with a D16Y5
I think I'm on the right track but I'm not sure of how it is finished; Since the equation representing the dissociation of HX is as follows; HX(aq) + H2O (l) <--> H3O+(aq) + X-(aq) It can then be shown that the equilibrium constant is found as such; Ka = [H3O+][X-]/[HX] (Note: H2O is not in the equilibrium equation because it is a liquid and further, the solvent) I first found [HX] as follows; [HX] = ((0.1246 M HX) x (25.00 mL) x (1L/1000 mL))/(40.12 mL x 1000 mL/1L) [HX] = 0.07764 M HX I then found the [H3O+] by converting the pH into [H3O+], given that pH = -log[H3O+] which results in; 5.9 = -log[H3O+] [H3O+] = 1.3x10-6 M Now as to how we solve for [X-] I'm not sure if [X-] = [H3O+] or what but if anyone can tell me if I'm completely off or somewhere near where I want to be perhaps they can finish it for you or give you the right answer
Icd 9 hx code vena cava filter
The dissociation constant is:k = [H][X]/[HX]
Honda HX has a CVT (continuously variable transmission).
hx
hx
You can change a Honda Civic HX transmission to a manual transmission. It is not an easy job to do.
1996-2000 civic lx and HX models, the LX is non v-tec and the HX is v-tec
Hx is an abbreviation for history.Hx means medical history on documents such as discharge papers and shot records.
The acid dissociation constant, Ka, is the ratio of the concentrations of the products (H+ and X-) to the concentration of the reactant (HX) at equilibrium. It is usually expressed as [H+][X-]/[HX]. The larger the Ka value, the stronger the acid.