Anything raised do 0 is always 1
I hope that’s correct
We find the first differences between terms:
7-4=3; 12-7=5; 19-12=7; 28-19=9.
Since these are different, this is not linear.
We now find the second differences:
5-3=2; 7-5=2; 9-7=2. Then:
Since these are the same, this sequence is quadratic.
We use (1/2a)n², where a is the second difference:
(1/2*2)n²=1n².
We now use the term number of each term for n:
4 is the 1st term; 1*1²=1.
7 is the 2nd term; 1*2²=4.
12 is the 3rd term; 1*3²=9.
19 is the 4th term; 1*4²=16.
28 is the 5th term: 1*5²=25.
Now we find the difference between the actual terms of the sequence and the numbers we just found:
4-1=3; 7-4=3; 12-9=3; 19-16=3; 28-25=3.
Since this is constant, the sequence is in the form (1/2a)n²+d;
in our case, 1n²+d, and since d=3, 1n²+3.
The correct answer is n²+3
Anyhoot, you can multiply 1.5 by 0.75 to find the time. And you can multiply by 60 to get 67.5 minutes, aka, 1.125 hours
Answer:
64k^2 - 16k +1
Step-by-step explanation:
We can rewrite this as
(-8k+1) ^2
We know that (a+b)^2 = a^2 +2ab +b^2
Let a = -8k and b = 1
(-8k+1) = (-8k)^2 +2*(-8k)(1) + 1^2
=64k^2 - 16k +1