To find perimeter you add up all the sides so the answer is 210
The vertical numbers are the distance and the horizontal numbers are the time in seconds.
Look at where the line is located at the number 20.
It si between the 2 and the 3 on the horizontal line, but you can see it is closer to the 2 than it is the 3, so the time would be 2.2s
Answer:
The 185th digit following that pattern of '12345678910111213141516' would be the number '5'
Step-by-step explanation:
The reason the 185th digit is '5' is because the pattern increases it's standard number by the following number (ex, 1,2,3,4,5, etc), and eventually it reached 185 as one of it's numbers, '5' would be the digit because the '18' part of 185 would only be the 184th digit, hence you are needed to include the '5' to complete the sequence of that number to reach the 185th digit.
The second question:
Consider the division expression 
. Select all multiplication equations that correspond to this division expression.
Answer:
1. See Explanation
2. and
Step-by-step explanation:
Solving (a):
Given
Required
Interpret 
in 2 ways
<u>Interpretation 1:</u> Number of groups if there are 5 students in each
<u>Interpretation 2:</u> Number of students in each group if there are 5 groups
<u>Solving the quotient</u>
<u>For Interpretation 1:</u>
The quotient means: 12 groups
<u>For Interpretation 2:</u>
The quotient means: 12 students
Solving (b):
Given
Required
Select all equivalent multiplication equations
Let ? be the quotient of t
So, we have:
Multiply through by 2
Rewrite as:
--- This is 1 equivalent expression
Apply commutative law of addition:
--- This is another equivalent expression
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
