ANSWER TO QUESTION 1
Method 1: Observing a pattern.
Perimeter is the distance around the triangle.
For the first triangle, the perimeter is

Let us write the sum of the 7s and 5s in such a way that we can easily recognize a pattern.

or

For the second triangle, the perimeter is ,


For the third rectangle the perimeter is


For the nth triangle the perimeter is,


This implies,

Method 2: Using the formula
We can write the perimeter as the sequence,

where the first term is

and the constant difference is

The nth term is given by the formula,



This simplifies to,

The correct answer is B.
ANSWER TO QUESTION 2
We examine the y-coordinates of the relation to see if there is a constant difference.

Since there is a constant difference, it means the relationship is linear.
To determine whether it is decreasing or increasing, we need to find the slope using any two points.

Since the slope is negative, the relationship is a function that is decreasing.
Therefore the function is decreasing and linear.
The correct answer is C.
ANSWER TO QUESTION 3
For the ordered pair


The relation between x and y is that,

For the ordered pair,


Their relation between x and y is,

For the ordered pair


The relation is

Also,


and finally,


In each case we raise 3 to the exponent of the x-value to get the y-value.
So in general, if we have

the rule will be

The correct answer is C.
ANSWER TO QUESTION 4
For the ordered pair,


This implies that,

we can rewrite in terms of the x-value to obtain,

For the ordered pair,


This implies that,

We can rewrite this in terms of x-value to get,

Similarly



Therefore the rule is

The correct answer is C.