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how can I solve this non linear partial differential equation mentioned below I have some initial & boundary condition
I need to obtain profile k^i =constant value of =6.54×10-3 μm -1 this is the boundary condition here x',y',z' are the coordi...
how can I solve this non linear partial differential equation mentioned below I have some initial & boundary condition
I need to obtain profile k^i =constant value of =6.54×10-3 μm -1 this is the boundary condition here x',y',z' are the coordi...
3년 초과 전 | 0
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how can I solve this non linear partial differential equation mentioned below I have some initial & boundary condition
거의 4년 전 | 답변 수: 2 | 0
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Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
대략 5년 전
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
대략 5년 전
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how to get color name as output corresponding to it vector as input
for eg if i give input vector a=[1,0,0] the output shold be color name corresponding to this vector
대략 5년 전 | 답변 수: 1 | 0
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Date Validation Logic and Error message
function isvalid = valid_date(y, m, d) % Check if the inputs are valid % Check that they are scalars if ~(isscalar(y) && iss...
Date Validation Logic and Error message
function isvalid = valid_date(y, m, d) % Check if the inputs are valid % Check that they are scalars if ~(isscalar(y) && iss...
5년 초과 전 | 0
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Write a function called minimax that takes M, a matrix input argument and returns mmr, a row vector containing the absolute values of the difference between the maximum and minimum valued elements in each row. As a second output argument called mmm,
function [mmr,mmm] = minimax(M) a=M'; mmr=max(a,[],1)-min(a,[],1); mmm= max(M(:)) - min(M(:)); end
Write a function called minimax that takes M, a matrix input argument and returns mmr, a row vector containing the absolute values of the difference between the maximum and minimum valued elements in each row. As a second output argument called mmm,
function [mmr,mmm] = minimax(M) a=M'; mmr=max(a,[],1)-min(a,[],1); mmm= max(M(:)) - min(M(:)); end
5년 초과 전 | 0
