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GarryVolchara [31]
3 years ago
8

If you create an object using Java’s BigDecimal class, and then perform a division that results in a non-terminating decimal div

ision such as 1/3, but specify that an exact result is needed, a(n) ____ is thrown.
Computers and Technology
1 answer:
azamat3 years ago
3 0

Answer:

Arithmetic Exception is the correct answer to the following blank.

Explanation:

Because the BigDecimal class is the class of the Java Programming Language that deal with the double data type numbers for the arithmetic expressions and also for the format conversions and it is the math type class of the Java Programming language which is used in arithmetic operations. So, that's why the following answer is not wrong.

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(a) Store last 7 digits of your student ID in a vector (7 element row or column vector). Write a MATLAB code which creates a 7x7
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Answer:

MATLAB code explained below with appropriate comments for better understanding

Explanation:

clc

clear all

ID = [1 2 3 4 5 6 7]; % Replace this with your student ID

%(a)

A = zeros(7);

for i=1:7

 

A(i,i)= ID(i);

 

end

fprintf('A =\n');

disp(A);

B = diag(ID);

fprintf('B =\n');

disp(B);

fprintf('Both A and B are same\n');

%(b)

if(mod(A(6,6),2)==0)

fprintf('A(6,6) is even\n');

else

fprintf('A(6,6) is odd\n');

end

%(c)

if(A(3,3)>0)

fprintf('A(3,3) is positive\n');

else if(A(3,3)<0)

fprintf('A(3,3) is negative\n');

else

fprintf('A(3,3)=0\n');

end

end

%(d)

fprintf('\nRequired series : ');

n = 35;

while n>=0

fprintf('%i',n);

if(n>0)

fprintf(', ');

end

n = n - 5;

end

fprintf('\n');

%(e)

n = input('\nInput an integer : ');

fprintf('%i! = ',n);

F = 1;

while n>1

F = F * n;

n = n - 1;

end

fprintf('%i\n',F);

%(f)

clear all

x = [3 7 2 1];

y = [4 3 9 1];

A = 0;

C = x(1);

for i=1:length(x)

A = A + x(i)*y(i);

B(i) = x(i)/y(i);

if(min(x(i),y(i))<C)

C = min(x(i),y(i));

end

end

C = 1/C;

fprintf('\nA = %i\n',A);

fprintf('B = ');

disp(B)

fprintf('C = %0.5g\n',C);

%(g)

clear all

A = randi([5 25],[1,10]);

maxA = A(1);

for i = 2:10

if(maxA<A(i))

maxA = A(i);

end

end

minA = A(1);

i = 2;

while i<11

if(minA>A(i))

minA = A(i);

end

i = i+1;

end

fprintf('\nA = ');

disp(A);

fprintf('maxA = %i\n',maxA);

fprintf('minA = %i\n',minA);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%output

A =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

B =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

Both A and B are same

A(6,6) is even

A(3,3) is positive

Required series : 35, 30, 25, 20, 15, 10, 5, 0

Input an integer : 6

6! = 720

A = 52

B = 0.7500 2.3333 0.2222 1.0000

C = 1

A = 25 14 7 10 13 17 10 17 19 9

maxA = 25

minA = 7

>>

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