If Beck has 5 pairs of shoes, that means he has 10 shoes because there are 2 shoes in a pair.

\(\displaystyle 5\times2=10\)

Since each shoe has 6 buttons, that means that there are 60 buttons in the collection. 

\(\displaystyle 10\times6=60\)

Simplify the numerator and denominator by cancelling out common factors.  

\(\displaystyle \left(-\frac{3}{8}\right)\left(\frac{2}{7}\right)\)

Then follow suit and multiply the denominator with denominator and numerator with numerator.

\(\displaystyle =\left(-\frac{3}{4}\right)\left(\frac{1}{7}\right) = -\frac{3}{28}\)

Multiply the ones digits together.

\(\displaystyle 9\times 7 =63\)

The ones digit of the final answer is \(\displaystyle 3\).

Carry over the \(\displaystyle 6\) to the next calculuation.

Multiply the \(\displaystyle 7\) with the tens digit of the first number, and add the carryover \(\displaystyle 6\).

\(\displaystyle 7\times 3 +6 = 21+6 = 27\)

The answer is:  \(\displaystyle 273\)

Eliminate the absolute value sign before proceeding to multiply all the terms.  

A value inside the absolute value will result into a positive value.

\(\displaystyle (-4)(9)\left | -9\right | = (-4)(9)(9)\)

From here, first multiply \(\displaystyle -4\cdot 9\), when doing so remember when a negative number is multiplied by a positive number the resulting product is negative.

Therefore we get,

\(\displaystyle -4\cdot 9=-36\).

Now we will need to multiply \(\displaystyle -36\) with \(\displaystyle 9\).

First multiply six with nine to get \(\displaystyle 54\). Keep the four in the ones place and carry the five to the tens place. Now multply nine with three \(\displaystyle 9\cdot 3=27\). Since we carried the five over we need to add \(\displaystyle 5+27=32\). Combine this value with the value that was found for the ones spot to get, \(\displaystyle 324\). Finally, place a negative sign in front of it to arrive at the final answer.

\(\displaystyle -324\).

Multiply the ones digits.

\(\displaystyle 3\times 8 = 24\)

The \(\displaystyle 4\) is the ones digit of the final answer. The \(\displaystyle 2\) will be the carryover for the next calculation.

Multiply the tens digit of \(\displaystyle 13\) with \(\displaystyle 8\) with the carryover.

\(\displaystyle 1\times 8+2 = 10\)

There are no further calculuations. Combine the numbers.

The answer is:  \(\displaystyle 104\)

Multiply the ones digit of \(\displaystyle 321\) with \(\displaystyle 9\).

\(\displaystyle 1\times 9=9\)

Multiply the tens digit of \(\displaystyle 321\) with \(\displaystyle 9\).

\(\displaystyle 2\times 9 =18\)

Since this number is 10 or greater, use this tens digit as the carry over for the next calculation.

Multiply the hundreds digit of \(\displaystyle 321\) with \(\displaystyle 9\) with the carry over.

\(\displaystyle 3\times 9+1 = 28\)

Combine this number with the ones digit of the previous calculations.

The correct answer is:  \(\displaystyle 2889\)

The numbers are positive. We just multiply. Answer is \(\displaystyle 42\). 

There is a negative number and zero. Regardless whether the number is positive or negative, anything multiplied by zero is always zero. Answer is \(\displaystyle 0\).

We have one positive and one negative number.

When multipled, our answer is negative.

The product is \(\displaystyle -36\).

We have two negative numbers. When multiplied, the answer is positive.

\(\displaystyle -15\\ \underline{\times -11}\)

          \(\displaystyle 15\)

    \(\displaystyle \underline{+150}\)

        \(\displaystyle 165\)

The product is \(\displaystyle 165\).