Answer:
The first term is when n= 1
3(1)+2 = 3+2
= 5
The second term is when n= 2
3(2)+2 = 6+2
= 8
The third term is when n=3
3(3)+2 = 11
Step-by-step explanation:
To find the gradient:
y2-y1/x2-x1
Pick any two coordinates
11)
(4,3) and (1,-1)
-1-3= -4
1-4= -3
-4/-3 = 1.3(1dp)
12)
(0,1) and ( 2,-2)
-2-1= -3
2-0= 2
-3/2= -1.5
Might be wrong but this is what I got from what I have learnt.
Hope this helps! :)
The first limit is known as a left-hand limit. We're approaching x = 3 from the left. This means we start with values smaller than x = 3, say x = 2.5 and move closer to x = 3.
Because x is starting off smaller than 3, we use the first piece of the piecewise function. This is the piece that corresponds to x < 3. So we plug x = 3 into that piece to get
2x^2 - x = 2(3)^2 - 3 = 2(9) - 3 = 18 - 3 = 15
So the result for the left-hand limit is 15.
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The right hand limit will have us start on the other side of x = 3. We start at x = 3.5 and move closer to x = 3. So we'll use the

portion.
Plug x = 3 into the second piece to get
3 - x = 3 - 3 = 0
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The result of the left-hand limit was 15
The result of the right-hand limit was 0
The fact that the results do not match up means that the overall limit at x = 3 does NOT exist.
Step-by-step explanation:
total outcomes = 720
favourable outcomes =8
=8/720
=0.01111