i want to Write a code or script including a FOR LOOP in order to computing the value of d for the following values of x and returning an output variable named ANSWER just as shown : x = 0.10, x = 0.15, and x = 0.20

3 comentarios

Daniel Pollard
Daniel Pollard el 15 de Abr. de 2021
Can you give more detail, such us -
  1. Where do these numbers come from? How were they found in the first place?
  2. What have you tried so far?
  3. What sort of calculation do you expect to be inside the for loop?
Hamada Alkhlif
Hamada Alkhlif el 15 de Abr. de 2021
sorry i forgot to entionb the eqation that we should use for d
  1. d=((34.63/x)-5.162)/2.54
  1. d = [];
  2. for x=[0.1000,0.1500,0.2000]
  3. d=[d ((34.63/x)-5.126)/2.54];
  4. disp ("ANSWER");
  5. end
  6. x=[0.1000 0.1500 0.2000];
  7. fprintf("\t%4g\t\t%4g\n",[x;d])
but when i put this cod into matlab it display like this
for x it shoulkd be 4 decimals .
DGM
DGM el 15 de Abr. de 2021
Editada: DGM el 15 de Abr. de 2021
Try
fprintf("\t%8.4f\t%8.4f\n",[x;d])
using %g strips insignificant trailing zeros

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 Respuesta aceptada

Daniel Pollard
Daniel Pollard el 15 de Abr. de 2021
Editada: Daniel Pollard el 15 de Abr. de 2021

0 votos

Your code is
d = [];
for x=[0.1000,0.1500,0.2000]
d=[d ((34.63/x)-5.126)/2.54];
disp ("ANSWER");
end
x=[0.1000 0.1500 0.2000];
fprintf("\t%4g\t\t%4g\n",[x;d])
If I understand right, you want
d = [];
x=[0.1000,0.1500,0.2000];
for xi = 1:numel(x)
d=[d ((34.63/x(xi))-5.126)/2.54];
disp ("ANSWER");
fprintf("\t%5.4f\t\t%.4f\n", [x(xi);d(xi)])
end

7 comentarios

Hamada Alkhlif
Hamada Alkhlif el 15 de Abr. de 2021
answer for your first code
>> Untitled6
ANSWER
ANSWER
ANSWER
0.1 134.32
0.15 88.8743
0.2 66.1512
>>
answer for second code you gave :
>> Untitled3
ANSWER
0.1000 134.3205
ANSWER
0.1500 88.8743
ANSWER
0.2000 66.1512
>> this one coorect accept the fact that there are 3 answer words. it must be 1 answer word and 3 rows and 2 coloms with 4 decimals for x' and d'
Daniel Pollard
Daniel Pollard el 15 de Abr. de 2021
My apologies - try
d = [];
x=[0.1000,0.1500,0.2000];
disp ("ANSWER");
for xi = 1:numel(x)
d=[d ((34.63/x(xi))-5.126)/2.54];
fprintf("\t%5.4f\t\t%.4f\n", [x(xi);d(xi)])
end
Hamada Alkhlif
Hamada Alkhlif el 15 de Abr. de 2021
Editada: Hamada Alkhlif el 15 de Abr. de 2021
almost the correct answer after i use the code in matlab i got this answer :
ANSWER
0.1000 134.3205
0.1500 88.8743
0.2000 66.1512
which is almost the same as the output asked above but one thing does not seems right here ,the last 3 numbers of (d') are not alighned , in other word (5 ,3 .2 ) should be alighned vertically .
thank you @Daniel Pollard
DGM
DGM el 15 de Abr. de 2021
Use a wider field width as in the example I gave
fprintf("\t%8.4f\t%8.4f\n",[x;d])
It could be tightened up as necessary.
Hamada Alkhlif
Hamada Alkhlif el 15 de Abr. de 2021
i did and finally i got the wanted answer .
thank you @DGM
Hamada Alkhlif
Hamada Alkhlif el 15 de Abr. de 2021
thank you @Daniel Pollard
Hamada Alkhlif
Hamada Alkhlif el 15 de Abr. de 2021
thanks everybody

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Más respuestas (1)

Jan
Jan el 15 de Abr. de 2021
Editada: Jan el 15 de Abr. de 2021

0 votos

disp ("ANSWER");
for x = [0.10, 0.15, 0.20]
d = ((34.63 / x) - 5.126) / 2.54;
fprintf("%12g%12g\n", x, d)
end
Or:
x = [0.10, 0.15, 0.20]
d = ((34.63 ./ x) - 5.126) / 2.54; % .7 for elementwise division
fprintf('Answer:\n');
fprintf("%12g%12g\n", [x, d].')

1 comentario

Hamada Alkhlif
Hamada Alkhlif el 15 de Abr. de 2021
1st code gives :
>> Untitled3
ANSWER
0.1 134.32
0.15 88.8743
0.2 66.1512
>> the answer here does not have 4 decimals for x' and not alighed vertically , same for d'
2nd code gives :
>> Untitled3
x =
0.1000 0.1500 0.2000
Answer:
0.1 0.15
0.2 134.32
88.8743 66.1512
>>

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el 15 de Abr. de 2021

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Jan
el 15 de Abr. de 2021

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