Examples of this are:
2 and 
2 and 
4 and 
I'm guessing you have a work sheet going along with this which has actual images of containers. I've attached an example - if you can post the actual work sheet you're referring to I can edit this answer to correctly reflect your specific question.
The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
<u>Property of rhombus:
</u>
Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.
so if you do 9 divided by 60 and it will give you 0.15 and that is your answer
The first one is a triangle because: The two smaller sides of a triangle must add up to be larger than the longest side.
14 + 12 = 26 > 19
ANSWER
EXPLANATION
We want to determine the value of f(3) that will lead to an average rate of change of 19 over the interval [3, 5].
The average rate of change of f(x) over the interval [a,b]:
If the average rate of change over the interval [3, 5] is 19, then;
From the to table f(5)=13
