Recent content by Punchlinegirl

Ok so if I did this right, we have 9.12e-8 mol/L of H+ ions. Then we had 100.09 mols of CaCO3. I also know that the pH of the solution after neutralization is 4.36, but I don't really understand how to plug these into the equation since all of the units are different.

Homework Statement Estimate the volume of acid rain in L that can be neutralized by 1.00 g of limestone. Assume the pH of rainwater collected in your rain gauge is a representative value for acid rain. Assume that acid rain contains sulfuric acid as the source of hydrogen ions. The molecular...

Homework Statement Suppose that we are testing H_0: \mu =\mu_0 versus H_1: \mu \neq\mu_0. Calculate the P-value for the following observed values of the test statistic. a: z_0 =2.45 e: z_0=-0.25 Homework Equations none The Attempt at a Solution I got part a by using the...

A mass weighing 8 lb stretches a spring 1.5 in. The mass is also attached to a damper with coefficient γ . Determine the value of γ for which the system is critically damped; be sure to give units for γ . I have no idea where to start. Can someone help me out?

Oh sorry, I meant r= 1 then my homogenous solution would be c_1e^t +c_2te^t

Find the solution of the given inital value problem. y"-2y'+y=te^t+4 y(0)=1 y'(0)=1 r^2-2r+1=0 (r-1)^2 so r= -1 my homogenous solution is y_h(t)=c_1e^-t+c_2te^-t I have no idea where to begin for the particular solution. I know a need a guess, and I would think it...

Find the solution of the given initial value problem y'-y=2te^2t y(0)=1 so p(t)=-1 then \mu (t)= e^/int -1 dt= e^-t e^-t y'-e^-t y=2te^2te^-t d/dt [e^-t y]=2te^t dt e^-t = \int 2te^t dt e^-t y= 2te^t-2e^t y=-2(t-1)+ce^-t plugging in the initial conditon gives me...

Use reduction of order to find a second solution to the given differential equation (x-1)y"-xy'+y=0 x>1 y_1=e^x Putting it in standard form gives y"-x/x-1 y' + 1/x-1 y =0 y(t)=v(t)e^x y'(t)= v'(t)e^x +v(t)e^x y"(t)= v"(t)+2v'(t)e^x +v(t)e^x plugging into...

Use reduction of order to find a second solution of the given differential equation t^2y"-4ty'+6y=0 t>0 y_1(t)=t^2 first I put in in the standard form y"-4/ty'+6/t^2y =0 then y= v(t)t^2 y'=v'(t)t^2+v(t)2t y"=y"(t)t^2+v'(t)2t+v'(t)2t+v(t)2 = t^2v"(t)+4tv'(t)+2v(t)...

I get it.. thanks a lot!

Solve the given initial value problem. 9y"-12y'+4y=0 y(0)=2 y'(0)=-1 9r^2-12r +4=0 (3r-2)^2 so r=2/3 so my general solution would be y(t)=c_1e^2/3t +c_2e^2/3t Whenever I try to use the initial conditions I can't get them to work. One disappears. Is my general...

so to find the roots I'd use the quadratic formula -2 +/- \sqrt 4-8 /4 so then would my roots be -.5 +/- 2i?

I'm feeling really stupid right now, but I can't seem to get this to factor. 2r^2 +2r+1= 0 Can someone please help me out?

Draw a direction field for the given differential equation. Based on the the direction field, determine the behavior of y at t goes to infinity. If this behavior depends on the initial value of y at t=0, describe the dependency. y'= -1-2y In class we did examples where we had a range...

I cross multiplied and got, (6.4e-11)(2.25+3y+y^2) = 2.96e-10(.25+y+y^2) 1.44e-10+1.92e-10y+6.4e-11y^2= 7.4e-11+2.96e-10y +2.96e-10y^2 Then my quadratic came out to be -2.32e-10y^2 -1.04e-10y + 7e-11= 0Then by the quadratic formula I got .369