\(\displaystyle 18\div(5-2)\)

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the division. 

\(\displaystyle 5-2=3\)

\(\displaystyle 18\div3=6\)

\(\displaystyle 16+3\times9+(42-10)\)

When solving this problem, remember order of operations PEMDAS. The parentheses come first, followed by the multiplication, and then addition. 

\(\displaystyle 42-10=32\)

\(\displaystyle 3\times9=27\)

\(\displaystyle 16+27+32=75\)

In ten years, Mark will be \(\displaystyle 43\) years old, so Mark is \(\displaystyle 43-10 = 33\) years old now, and Brian is one-third of this, or \(\displaystyle 33 \div 3 = 11\) years old. 

Let \(\displaystyle N\) be the number of years in which Mark will be twice Brian's age. Then Brian will be \(\displaystyle N + 11\), and Mark will be \(\displaystyle N + 33\). Since Mark will be twice Brian's age, we can set up and solve the equation:

\(\displaystyle 2 (N + 11) = N + 33\)

\(\displaystyle 2N + 22 = N + 33\)

\(\displaystyle 2N + 22-N - 22 = N + 33 -N - 22\)

\(\displaystyle N = 11\)

Mark will be twice Brian's age in \(\displaystyle 11\) years.

Since Gary will be 37 in five years, he is \(\displaystyle 37 - 5 = 32\) years old now. He is twice as old as Cathy, so she is \(\displaystyle 32 \div 2 =16\) years old, and in five years, she will be \(\displaystyle 16 + 5 = 21\) years old.

Let \(\displaystyle X\) be the number of hours that Cliff parks his car in the garage. Since he pays a flat fee of \(\displaystyle \$7\) plus \(\displaystyle \$2\) per hour or portion thereof, he pays \(\displaystyle 2X + 7\) dollars. Since he does not want to spend more than \(\displaystyle \$20\),

\(\displaystyle 2X + 7 \leq 20\)

\(\displaystyle 2X + 7 - 7 \leq 20 - 7\)

\(\displaystyle 2X \leq 13\)

\(\displaystyle 2X \div 2 \leq 13 \div 2\)

\(\displaystyle X \leq 6.5\)

Since a portion of an hour counts as much as an hour, \(\displaystyle X\) must be a whole number.

\(\displaystyle X \leq 6.5\)

implies that

\(\displaystyle X \leq 6\).

Cliff can park in the garage for a maximum of 6 hours, and since he entered the garage at \(\displaystyle 7:00\textup { AM}\), he must exit \(\displaystyle 6\) hours later:

\(\displaystyle 7+ 6 - 12 = 1\),

or \(\displaystyle 1:00\textup { PM}\).

\(\displaystyle (12+13)\times2\)

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication. 

\(\displaystyle 12+13=25\)

\(\displaystyle 25\times2=50\)

\(\displaystyle (34+17)\times2\)

When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication. 

\(\displaystyle 34+17=51\)

\(\displaystyle 51\times2=102\)

If \(\displaystyle B\) is odd, then \(\displaystyle AB\) is odd, since the product of two odd whole numbers must be odd. When the odd number \(\displaystyle C\) is added, the result, \(\displaystyle AB +C\), is even, since the sum of two odd numbers must be even.

If \(\displaystyle B\) is even, then \(\displaystyle AB\) is even, since the product of an odd number and an even number must be even. When the odd number \(\displaystyle C\) is added, the result, \(\displaystyle AB +C\), is odd, since the sum of an odd number and an even number must be odd.

Combine all the like terms.

The \(\displaystyle x^2\) terms can be combined together, which gives you \(\displaystyle 13x^2\).

When you combine the \(\displaystyle x\) terms together, you get \(\displaystyle -2x\).

There is only one \(\displaystyle y\) term so it doesn't get combined with anything. Put them all together and you get 

\(\displaystyle 13x^2-2x+5y\).

\(\displaystyle \dpi{100} \small 2(4x-3x)-6t+5x\)

First distribute the 2:    \(\displaystyle \dpi{100} \small 8x-6x-6t+5x\)

Combine the like terms:      \(\displaystyle \dpi{100} \small 7x-6t\)