Answer:
Word problem 3
Step-by-step explanation:
Word Problem one is a subtraction problem since Felicia is giving away two marbles to Lucas. Word Problem two is a division problem since Felicia is evenly dividing the number of marbles she has into two bags. Word Problem three is a multiplication problem because Ryan has two times as many marbles that Felicia does. Word Problem four is an addition problem because she found two more marbles while she already has eight marbles. Keywords for subtraction word problems are: fewer than, decrease, take away, less than, minus, difference, change, lost reduced, and subtract. Keywords for division word problems are: As much, cut up, groups, equally, sharing, half, how many in each, parts, per percent, quotient, ratio, and separated. Keywords for multiplication word problems are: double, every, factor, increased, multiplied product, times, and tripled.
Sum/difference:
Let

This means that

Now, assume that
is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that
was rational, which proves that the sum/difference of the two given terms was irrational
Multiplication/division:
The logic is actually the same: if we multiply the two terms we get

if again we assume x to be rational, we have

But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.
We cant get 2 fours in a single die.we need at least 2 die to get 2 fours.
Section J has 20 rows....each row has 15 seats....so section J has (20 * 15) = 300 seats. If each seat costs $ 18, and all the seats are sold, then the amount of money collected would be : (300 * 18) = $ 5,400 <=