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sfolax6776
Jul 10, 2018

5.11 What are the multiple phases of programming?

5 comments

Edited: Nov 16

A. Introduction:

When you try to solve a problem like the one below, you have to come up with the algorithm that yields the area first, then code the algorithm. Can you skip the step of algorithm design?

 

B. Watch(Slides)

C. Try: R6.17

R6.17 Write pseudocode for a program that reads a sequence of student records and prints
the total score for each student. Each record has the student’s first and last name,
followed by a sequence of test scores and a sentinel of –1. The sequence is terminated
by the word END. Here is a sample sequence:
Harry Morgan 94 71 86 95 -1
Sally Lin 99 98 100 95 90 -1
END
Provide a trace table for this sample input.

 

rtptyu
Jul 15, 2018

d/2=6 c/2=5

6^2-5^2=(d-b)^2

6- √ 11=b

b=2.7 b=d-√ 11=d-2,7

A=2/3c(d-2,7)+(d-2,7)^3/2c

0
mr.ricklu
Jul 15, 2018

When d=12 and c=10

=0.9660824973853709

 

double d=?;

double c=?;

double h= (d/2)- Math.sqrt(Math.pow(d/2, 2)-Math.pow(c/2, 2));

System.out.println((2/3*c*h)+(Math.pow(h, 3)/(2*c)));

 

0
sfolax6776
Jul 15, 2018

Great job, can you guys develop a program to calculate the area and upload your complete code here?

0
wangernestrdragon
Jul 15, 2018

double d= 12;

double c= 10;

double h= (12+ Math.sqrt(Math.pow(12,2)-100))/2;

double A= (2/3*c*h)+(Math.pow(h, 3)/(2*c));

System.out.println(A);

 

ans=40.43391750261463

0
tian
Jul 15, 2018

double d = 12;

double c = 10;

double h = d/2-Math.sqrt((d/2)*(d/2) - (c/2) * (c/2));

 

System.out.print((2*c*h/3)+((Math.pow(h, 3))/(2*c)));

0
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