# LissD - %Shear Rate Data(1/s g gdot=[0.91 3.3 4.1 6.3 9.6...

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%Shear Rate Data (1/s) gdot=[0.91 3.3 4.1 6.3 9.6 23 36 49 65 105 126 215 315 402]; g %Shear Stress Data (N/m^2) tau=[0.059 0.15 0.19 0.27 0.39 0.87 1.33 1.65 2.11 3.44 4.12 7.02 10.21 13.01]; t %code to display the data % %Linear Regression for the casson values cassongdot = [.91 3.3 4.1 6.3 9.6 23 36 49]; cassontau = [.059 .15 .19 .27 .39 .87 1.33 1.65]; c cassony = cassontau.^(1/2); cassonx = cassongdot.^(1/2); c %This is where I found my C matrix to do the linear regression sumcassongdot = sum (cassonx); sumcassongdot2 = sum (cassonx.^2); sumcassontaugdot = sum (cassonx * cassony'); sumcassontau = sum(cassony); s %D is the C matrix and O is the B matrix in the equation Ca=B D = [ 8 sumcassongdot ; sumcassongdot sumcassongdot2]; O = [ sumcassontau ; sumcassontaugdot]; O Acasson = inv(D)*O; Constantcasson = Acasson(2,1); tauycasson = Acasson(1,1)^2; Srcasson = 0; iD = inv(D); i %I used a for loop to find the standard deviaton. for n = 1:1:8

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LissD - %Shear Rate Data(1/s g gdot=[0.91 3.3 4.1 6.3 9.6...

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