Ignore what I have written because Im pretty sure its wrong. Thank.pdf

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Ignore what I have written because I\'m pretty sure its wrong. Thanks! Deliver the assignment to my office, 4502 French Hal! West or Scan your assignment and send it as an email attachment. Write the complex number, z = a + bi, in polar form, z = r (cos theta + i sin theta), with argument theta between 0 and 2 pi. -3i 3 - Squareroot 3i -1 + i II. Find the product, z_1z_2, and the quotient, z_1/z_2. Express your answer in polar form. z_1 = 3(cos 15 degree + i sin 15 degree) z_2 = 6(cos 20 degree + i sin 20 degree) z_1z_2: ____ z_1/z_2: ____ z_1 = 21 (cso pi/5 + i sin pi/5) z_2 = 7(cos 2 pi/3 + i sin 2 pi/3) z_1z_2: ____ z_1/z_2: ____ Solution I) 1) -3i r=[02+(-3)2]=3 =tan-1(-3/0) =3/2 polar form is 3(cos(3/2)+i sin(3/2)) 2) 3-3i r=[32+(-3)2]=12=23 =tan-1(-3/3) =11/6 polar form is 23(cos(11/6)+i sin(11/6)) 3) -1+i r=[(-1)2+12]=2 =tan-1(1/-1) =3/4 polar form is 2(cos(3/4)+i sin(3/4)) =========================================================== II) 1) z1=3(cos15o+isin15o), z2=6(cos20o+isin20o) z1z2=3(cos15o+isin15o)*6(cos20o+isin20o) z1z2=3*6(cos(15+20)o+isin(15+20)o) z1z2=18(cos(35)o+isin(35)o) z1/z2=3(cos15o+isin15o)/[6(cos20o+isin20o)] z1/z2=(3/6)(cos(15-20)o+isin(15-20)o) z1/z2=(1/2)(cos(-5)o+isin(-5)o) z1/z2=(1/2)(cos(355)o+isin(355)o) 2) z1=21(cos(/5)+isin(/5)), z2=7(cos(2/3)+isin(2/3)) z1z2=21(cos(/5)+isin(/5))*7(cos(2/3)+isin(2/3)) z1z2=21*7(cos((/5)+(2/3))+isin((/5)+(2/3))) z1z2=147(cos(13/15)+isin(13/15)) z1/z2=21(cos(/5)+isin(/5))/7(cos(2/3)+isin(2/3)) z1/z2=3(cos((/5)-(2/3))+isin((/5)-(2/3))) z1/z2=3(cos(-7/15)+isin(-7/15)) z1/z2=3(cos(23/15)+isin(23/15)).

Ignore what I have written because I'm pretty sure its wrong. Thanks! Deliver the assignment to
my office, 4502 French Hal! West or Scan your assignment and send it as an email attachment.
Write the complex number, z = a + bi, in polar form, z = r (cos theta + i sin theta), with argument
theta between 0 and 2 pi. -3i 3 - Squareroot 3i -1 + i II. Find the product, z_1z_2, and the
quotient, z_1/z_2. Express your answer in polar form. z_1 = 3(cos 15 degree + i sin 15 degree)
z_2 = 6(cos 20 degree + i sin 20 degree) z_1z_2: ____ z_1/z_2: ____ z_1 = 21 (cso pi/5 + i sin
pi/5) z_2 = 7(cos 2 pi/3 + i sin 2 pi/3) z_1z_2: ____ z_1/z_2: ____
Solution
I)
1)
-3i
r=[02+(-3)2]=3
=tan-1(-3/0) =3/2
polar form is 3(cos(3/2)+i sin(3/2))
2)
3-3i
r=[32+(-3)2]=12=23
=tan-1(-3/3) =11/6
polar form is 23(cos(11/6)+i sin(11/6))
3)
-1+i
r=[(-1)2+12]=2
=tan-1(1/-1) =3/4
polar form is 2(cos(3/4)+i sin(3/4))
===========================================================
II)
1)
z1=3(cos15o+isin15o), z2=6(cos20o+isin20o)
z1z2=3(cos15o+isin15o)*6(cos20o+isin20o)
z1z2=3*6(cos(15+20)o+isin(15+20)o)
z1z2=18(cos(35)o+isin(35)o)
z1/z2=3(cos15o+isin15o)/[6(cos20o+isin20o)]
z1/z2=(3/6)(cos(15-20)o+isin(15-20)o)

z1/z2=(1/2)(cos(-5)o+isin(-5)o)
z1/z2=(1/2)(cos(355)o+isin(355)o)
2)
z1=21(cos(/5)+isin(/5)), z2=7(cos(2/3)+isin(2/3))
z1z2=21(cos(/5)+isin(/5))*7(cos(2/3)+isin(2/3))
z1z2=21*7(cos((/5)+(2/3))+isin((/5)+(2/3)))
z1z2=147(cos(13/15)+isin(13/15))
z1/z2=21(cos(/5)+isin(/5))/7(cos(2/3)+isin(2/3))
z1/z2=3(cos((/5)-(2/3))+isin((/5)-(2/3)))
z1/z2=3(cos(-7/15)+isin(-7/15))
z1/z2=3(cos(23/15)+isin(23/15))

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Background: Sometimes the standard C libraries (stdio.h, stdlib.h, etc) are not extensive enough to handle every programming need. If you find yourself rewriting or reusing functions you’ve written earlier like those used to calculate statistical parameters (mean, variance etc), copying the code again and again is inefficient. If you could write the code once, troubleshoot it so it works, and have it compiled in its own object file, you could then link these object files to your source code object file and not have to rewrite all the functions. ------------------------------------------------ --------------------------------------------------------------------------------------- Page 2 Part I – [100 points] Description Your task is to create a personal library to implement a statistic package on any given data (assume that the data is always of type float). To do this you must create: 1. a header file (stats.h) with 5 function prototypes: § maximum() § minimum() § mean() – equation: = = 1 0 [ ] 1 N i x i N x § variance() – equation: 2 1 0 2 ( [ ] ) 1 1 x i x N N i = = § histogram() 2. five (5) implementation files with each of functions listed above – min.c, max.c, mean.c, variance.c, and histogram.c You will then write C code (main.c) that asks the user for the data file he/she wishes to compute the statistics for and displays the results to the user. YOU WILL NOT WRITE ANY STATISTICS FUNCTIONS IN main.c – you will simply call the functions prototyped in stats.h and coded in min.c, max.c, mean.c, variance.c, and histogram.c!!! Procedure: (Refer to textbook pg 669-677) 1. Create the header file stats.h: a. Start with a block comment summarizing the purpose of the header file b. # define any constant that you need for the header file (NOT YOUR CODE) c. Individual function prototypes (with introductory comments stating purpose of each function) d. Save e. Below is the template for stats.h #ifndef STATS_H #define STATS_H float minimum(…); // NOTE: You need to complete the prototypes float maximum(…); float mean(…); float variance(…); void histogram(…); #endif // STATS_H_INCLUDED EGR 107 Page 3 2. Create 5 implementation files (min.c, max.c, mean.c, variance.c, histogram.c) a. Start with a block comment summarizing the purpose of each implementation file b. #include the necessary standard C libraries for each function c. #include “stats.h” – the angular brackets (< >) are used for standard C libraries found in the system directory, the quotation marks (“ ”) indicate a custom library found in the SAME directory as your code. d. #define – any constants for that particular implementation file e. Each file will contain one function as you would normally write it. i. The first four files will calculate maximum, minimum, mean, and variance and return one value each time (so they can be regular call-by-value functions) ii. Your histogram function will count the number of occurrences of data within a certain range; each range is commonly referred to as a.

Background Sometimes the standard C libraries (stdio.h, stdlib.h, e.pdf

Background Sometimes the standard C libraries (stdio.h, stdlib.h, e.pdf

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ebrahimbadushata00

a. Modify the C program ex.9 so that it simulates the Unix pipe command: ls | grep .c | tail -4 b.. Implement following programs to exhibit UNIX Process Control: A program where parent process sorts array elements in descending order and prints them out and child process sorts array elements in ascending order and prints them out. A program where parent process counts number of vowels in the given sentence and child process will count number of words in the same sentence. The above programs should use system calls like fork, exec and wait. ex.9: /* simulates unix pipe ls | grep .c */ #include #include #include #define RD 0 #define WT 1 int main() { int fd[2]; pipe(fd); switch(fork()) { // parent forks case -1: perror(\"Fork error\"); break; case 0: // child does the ls close(fileno(stdout)); // no need for stdout here dup(fd[WT]); // duplicate the descriptor for writing close(fd[WT]); close(fd[RD]); execlp(\"ls\", \"ls\", NULL); break; default: // parent does the grep close(fileno(stdin)); // no need for stdin here dup(fd[RD]); // duplicate the descriptor for reading close(fd[WT]); close(fd[RD]); execlp(\"grep\", \"grep\", \".c\", NULL); break; } return 0; } Solution #include #include #include #define RD 0 #define WT 1 int main() { int fd1[2]; int fd2[2]; pipe(fd1); switch(fork()) { case -1: perror(\"Fork error\"); break; case 0: // child does the ls close(fileno(stdout)); dup(fd[WT]); close(fd1[WT]); close(fd1[RD]); execlp(\"ls\", \"ls\", NULL); break; default: pipe(fd2); switch(fork()) { case -1: perror(\"Fork error\"); break; case 0: close(fileno(stdin)); dup(fd1[RD]); dup(fd2[RD]; close(fd1[WT]); close(fd1[RD]); close(fd2[WT]); close(fd2[RD]); execlp(\"grep\", \"grep\", \".c\", NULL); break; default: dup(fd2[RD]); close(fd2[WT]); close(fd2[RD]); execlp(\"tail\", \"tail\", \"-4\", NULL); break; } return 0; }.

a. Modify the C program ex.9 so that it simulates the Unix pipe comm.pdf

a. Modify the C program ex.9 so that it simulates the Unix pipe comm.pdf

a. Modify the C program ex.9 so that it simulates the Unix pipe comm.pdf

ebrahimbadushata00

A severe B12 deficiency can cause megaloblastic anemia but in severe .pdf

A severe B12 deficiency can cause megaloblastic anemia but in severe .pdf

A severe B12 deficiency can cause megaloblastic anemia but in severe .pdf

ebrahimbadushata00

When you are admitted to the hospital for surgery, you are very often administered intravenously a physiological saline solution (a solution of NaCl in water). Calculate the concentration of this solution knowing that the solution has to be isotonic with blood and knowing that the average osmotic pressure of blood is 7.7 atm at 25 oC. Solution Pi (osmotic pressure) = iMRT molarity M = Pi/iRT = 7.7 / 2x[0.082 x (273.15 +25)] = 0.16 M i = Hoff factor = 2 since NaCl dissociates as Na+ and Cl-..

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1. Ignore what I have written because I'm pretty sure its wrong. Thanks! Deliver the assignment to my office, 4502 French Hal! West or Scan your assignment and send it as an email attachment. Write the complex number, z = a + bi, in polar form, z = r (cos theta + i sin theta), with argument theta between 0 and 2 pi. -3i 3 - Squareroot 3i -1 + i II. Find the product, z_1z_2, and the quotient, z_1/z_2. Express your answer in polar form. z_1 = 3(cos 15 degree + i sin 15 degree) z_2 = 6(cos 20 degree + i sin 20 degree) z_1z_2: ____ z_1/z_2: ____ z_1 = 21 (cso pi/5 + i sin pi/5) z_2 = 7(cos 2 pi/3 + i sin 2 pi/3) z_1z_2: ____ z_1/z_2: ____ Solution I) 1) -3i r=[02+(-3)2]=3 =tan-1(-3/0) =3/2 polar form is 3(cos(3/2)+i sin(3/2)) 2) 3-3i r=[32+(-3)2]=12=23 =tan-1(-3/3) =11/6 polar form is 23(cos(11/6)+i sin(11/6)) 3) -1+i r=[(-1)2+12]=2 =tan-1(1/-1) =3/4 polar form is 2(cos(3/4)+i sin(3/4)) =========================================================== II) 1) z1=3(cos15o+isin15o), z2=6(cos20o+isin20o) z1z2=3(cos15o+isin15o)*6(cos20o+isin20o) z1z2=3*6(cos(15+20)o+isin(15+20)o) z1z2=18(cos(35)o+isin(35)o) z1/z2=3(cos15o+isin15o)/[6(cos20o+isin20o)] z1/z2=(3/6)(cos(15-20)o+isin(15-20)o)

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