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 $\displaystyle 165$.

We have two negative numbers and one positive number. We multiply from left to right. Two negative numbers multiplied make a positive number. Two positive numbers multiplied is also positive.

$\displaystyle -4\cdot -7=28$

Now the expression becomes,

$\displaystyle 28\cdot 9$

Answer is $\displaystyle 252$.