18. A shelf holds fewer than 50 cans. If all of the cans on this shelf were put into stacks of
five cans each, no cans would remain. If the
same cans were put into stacks of three cans
each, one can would remain. What is the
greatest number of cans that could be on
the shelf?

40

Step-by-step explanation:

Given that the number of cans are fewer than 50.

Let the number of cans =

It is given that no cans remain if they are put in the stacks of five cans each.

i.e. number of cans are a multiple of 5.

Let us have a look at the numbers in decreasing order which are multiples of 5 and are lesser than 50.

45, 40, 35, 30, ......

Now, let us check the numbers one by one.

If 45,

Now, they are placed on stacks of three each and one can remains.

So, the remainder when is divided by 3 will be 1.

But when 45 is divided by 3, remainder is 0.

Now, let us check next greatest number, which is 40.

When 40 is divided by 3, the remainder is 1.

Therefore, greater number of cans can be 40.

choice b is correct

step-by-step explanation:

we have been given the following system of linear inequalities;

we are required to graphically determine the solution to these system of inequalities. using a graphing tool, the solution is as shown in the attachment below;

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x