Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

A company distributes free candies to all students of x schools. Each school has (x+1) classes. The number of students in each class is 3 more than the number of classes in each school. Each student is given 4 candies. Part A) Write an expression to show the total number of candies distributed by the company in x schools. Part B) What would x(x+1) represent? When simplified, what would be the degree and classification of this expression? Part C) How can you calculate the total number of students in each school?

OpenStudy (akashdeepdeb):

No. of schools = x No. of classes in each school = x+1 No. of students in each class = 3 more than number of classes in each school. Thus, no. of students in each class = (x+1) + 3 = x+4 No. of candies per student = 4

OpenStudy (akashdeepdeb):

Part A) Write an expression to show the total number of candies distributed by the company in x schools. So the total number of candies is given 4 each to x+4 students in x+1 classes in x schools. So total number of candies = 4*(x+4)*(x+1)*x

OpenStudy (anonymous):

awesome thanks bro are u able to help me with part B and C

OpenStudy (akashdeepdeb):

Part B) What would x(x+1) represent? When simplified, what would be the degree and classification of this expression? x is the number schools x+1 is the number of classes in each school So x*(x+1) would represent the total number of classes all taken together. When simplified x(x+1) = \(x^2+1\) which is a second degree polynomial.

OpenStudy (akashdeepdeb):

I suppose, you can do the C part now. Try it, ask me if you have a problem.

OpenStudy (anonymous):

I cant figure out how to put this all together for an answer.

OpenStudy (anonymous):

Please stop copying and submitting to Ms Huebener...She is getting upset at all of us taking answers from here, she just wants us to learn

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!