It can penetrate the blood-brain barrier.
Lookup the isalpha() function in the C standard library, this function will check and return non-zero for alphabetic characters.
Answer:
Option d is the correct answer for the above question.
Explanation:
- The first loop of the program has a second loop and then the statement. In this scenario, the second loop executes for the value of the first loop and the statement executes for the value of the second loop.
- The first loop executes 4 times, Then the second loop or inner loop executes n times for the n iteration of the first loop, for example, 1 time for the first iteration of the first loop, 2 times for the second iteration of the first loop and so on.
- Then the inner loop executes (1+2+3+4) iteration which gives the result 10 iterations.
- The sum initial value is 0 and the "sum++", increase the value of the sum by 1.
- So the value of the sum becomes 10 after completing 10 iterations of the inner for loop.
- Hence the 10 will be the output. So the Option d is the correct answer while the other is not.
Answer:
1. PowerPoint online
2. Goógle Slides
Explanation:
There is a various free, proprietary, and online software for multimedia presentations available for use. Some of which include:
1. PowerPoint online: this is an online and free version of Microsoft Office PowerPoint software. Some of its features include text formatting, use of animations, cloud storage among others.
2. Goógle Slides: This is a product of Goógle to make the multimedia presentation of documents available online. It also offers many feature including cloud storage.
Answer:
PROGRAM QuadraticEquation
Solver
IMPLICIT NONE
REAL :: a, b, c
;
REA :: d
;
REAL :: root1, root2
;
//read in the coefficients a, b and c
READ(*,*) a, b, c
WRITE(*,*) 'a = ', a
WRITE(*,*) 'b = ', b
WRITE(*,*) 'c = ', c
WRITE(*,*)
// computing the square root of discriminant d
d = b*b - 4.0*a*c
IF (d >= 0.0) THEN //checking if it is solvable?
d = SQRT(d)
root1 = (-b + d)/(2.0*a) // first root
root2 = (-b - d)/(2.0*a) // second root
WRITE(*,*) 'Roots are ', root1, ' and ', root2
ELSE //complex roots
WRITE(*,*) 'There is no real roots!'
WRITE(*,*) 'Discriminant = ', d
END IF
END PROGRAM QuadraticEquationSolver
