Siddhartha S.

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MATLAB

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Question:

Implement histogram equalization on image using MATLAB.

Siddhartha S.

Answer:

clear all; close all; clc im = imread('cameraman.tif'); imhist = histeq(im); figure imshow(imhist); g = zeros(256,1); % find histogram of image for i = 1:size(im,1) for j = 1:size(im,2) g(im(i,j)+1) = g(im(i,j)+1) + 1; end end g = g/sum(g); % histogram equaliztion imnew = zeros(size(im,1),size(im,2)); for i = 1:size(im,1) for j = 1:size(im,2) imnew(i,j) = 256*sum( g(1:im(i,j) )); end end imnew = uint8(imnew); figure imshow(imnew); for i = 1:size(im,1) for j = 1:size(im,2) g(imnew(i,j)+1) = g(imnew(i,j)+1) + 1; end end

Calculus

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Question:

Find the slope of the tangent line to the graph xy=y^{2}*x^{2}+y+x+2 at the point (0,-2).

Siddhartha S.

Answer:

We can use implicit differentiation to find y' and then plug in the point (0,-2) to find the slope of the tangent line at that point. Taking the derivative gives y+xy'=2yy'x^{2}+y^{2}(2x)+y'+1, Plugging in (0,-2) gives -2+0=0+0+y'+1, or y'=-3, Thus, the slope of the tangent line to this graph at the point (0,-2) is -3. Since we have a point and the slope of the tangent line, we will use the point-slope form of the line. The equation of the tangent line at (0,-2) is y-(-2)=-3(x-0), or y+2=-3x,

Algebra

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Question:

If 3*x - y = 12. What is the value of 8^x/2^y ?

Siddhartha S.

Answer:

One approach is to express 8^x/2^y in the same base so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting 2^3 for 8 in the numerator of 8^x gives ( 2^3x) /2y Since the numerator and denominator of have a common base, this expression can be rewritten as 2^(3x−y). In the question, it states that 3x−y=12, so one can substitute 12 for the exponent, 3x−y. So the answer is 2^12 !

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