<h3>
Answer: (n-1)^2</h3>
This is because we have a list of perfect squares 0,1,4,9,...
We use n-1 in place of n because we're shifting things one spot to the left, since we start at 0 instead of 1.
In other words, if the answer was n^2, then the first term would be 1^2 = 1, the second term would be 2^2 = 4, and so on. But again, we started with 0^2 = 0, so that's why we need the n-1 shift.
You can confirm this is the case by plugging n = 1 into (n-1)^2 and you should find the result is 0^2 = 0. Similarly, if you tried n = 2, you should get 1^2 = 1, and so on. It appears you already wrote the answer when you wrote "Mark Scheme".
All of this only applies to sequence A.
side note: n is some positive whole number.
Hey there!!
Okay so here are the steps....
(5 + 2i) + (8 - 3i) <---- when we have exponents we should worry about the "i" because it is equal to √-1 but here it isn't the case.
<u>
(13 - i)</u><u /> <----- This is the answer!
<u />I hope this helps! ^__^
Answer:
Slope-int form: y = 3x+5
Standard form: y - 3x = 5.
Step-by-step explanation:
(reminder: slope-intercept form is expressed as y=mx+b, and standard form is expressed as ax+bx=c.)
Since the slope is 3, the coefficient of x is also 3, which makes the equation y=3x.
But the y coordinate of the equation at x = -2 is -6, so we need to add 5 to the end of the equation, leaving you with:
y=3x+5.
To convert it to standard form, subtract 3x from both sides:
y - 3x = 5.
I hope this helped you.
Answer:
Step-by-step explanation:
Given the first two numbers of a sequence as 2, 6...
If it is an arithmetic difference, the common difference will be d = 6-2 = 4
Formula for calculating nth term of an ARITHMETIC sequence Tn = a+(n-1)d
a is the first term = 2
d is the common difference = 4
n is the number if terms
Substituting the given values in the formula.
Nth term Tn = 2+(n-1)4
Tn = 2+4n-4
Tn = 4n-4+2
Tn = 4n-2
2) If the sequence us a geometric sequence
Nth term of the sequence Tn = ar^(n-1)
r is the common ratio
r is gotten by the ratio of the terms I.e
r = T2/T1
r = 6/2
r = 3
Since a = 2
Tn = 2(3)^(n-1)
Hence the nth term of the geometric sequence is Tn = 2(3)^(n-1)
1/32 maybe................
