Question

A company distributes free candies to all the 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. How can you calculate the total number of students in each school?

Answer 1

So we have X schools. Each school has X+1 classes. So we have \[x(x+1)=x^{2}+x\] classes.

Each class has 3 more students than the number of classes so: \[x^{2}+x+3\] students in each class for a total of \[(x^{2}+x)classes \times (x^{2}+x+3) \frac{students}{class}\] students.

This gives us:\[x^{4}+x^{3}+3x^{2}+x^{3}+x^{2}+3x=x^{4}+2x^{3}+4x^{2}+3x\] students.

Answer 2

@twvogels can you help me with anther question

Answer 3

Sure

Answer 4

Greg drew 3 squares with each side equal to z units. For each square, he does something different to it according to each part below.
Greg decreased the length and width of the square by t units each. What will be the change in the area of the square? Show your work

Answer 5

@twvogels

Answer 6

So for the large square the area is:
\[A_L=z^{2}\]
and for the small one it's:
\[A_S=(z-t)^{2}\]
Then we can subtract to get the difference.
\[A_L−A_S\]
\[z^{2}-(z-t)^{2}\]\[z^{2}-(z^{2}-2zt+t^{2})\]\[z^{2}-z^{2}+2zt-t^{2})\]\[2zt-t^{2}\]

Answer 7

WOW thank you so much you are amazing one more question if that is ok with you

Answer 8

Ok.

Answer 9

A: Divide (10x^4y^3 + 5x^3y^2 - 15x^2y - 25x^2y^4) by -5x^2y. Show your work.

B: How would your answer in Part A be affected if the x^2 variable in the denominator was just an x?

C:What is the degree and classification of the polynomial you got in Part A?

@twvogels

Answer 10

\[\frac{10x^{4}y^{3}+5x^{3}y^{2}-15x^{2}y-25x^{2}y^{4}}{-5x^{2}y}\]

Just to make sure I read it right.

Answer 11

yes

Answer 12

Ok, this is going to take some time. Hold on.

Answer 13

\[-(2x^{2}y^{2}+xy-5y^{3}-3)\]

Answer 14

wait which part is that the answer for is it for A

Answer 15

Yes, that's Part A.

Answer 16

Can you give me the answers for the other parts too

Answer 17

Ok, I'll need to work them out. Give me a minute.

Answer 18

B is just the same as a, multiplied by the -5 we removed:

\[5(2x^{2}y^{2}+xy−5y^{3}−3)\]

Answer 19

so \[5(2x^2y^2+xy-5y^3-3)\] is the answer

Answer 20

to part b

Answer 21

so the answer for c would be 3rd degree and a classification of 4 right @twvogels

Answer 22

Yes that's correct.

Answer 23

ok thank you so much i became a fan