Distances in 2- and 3-dimensions (and even higher dimensions) can be found using the Pythagorean theorem. The straight-line distance can be considered to be the hypotenuse of a right triangle whose sides are the horizontal and vertical differences between the coordinates.
Here, you have A = (0, 0) and B = (3, 6). The horizontal distance between the points is ...
... 3 - 0 = 3 . . . . the difference of x-coordinates
The vertical distance between the points is ...
... 6 - 0 = 6 . . . . the difference of y-coordinates
Then the straight-line distance (d) between the points is found from the Pythagorean theorem, which tells you ...
... d² = 3² + 6²
... d = √(9 + 36) = √45 ≈ 6.7 . . . units
Answer:
7.211
Step-by-step explanation:
-For two points in the complex plane, the distance between the points is the modulus of the of the difference of the two complex numbers.
-Point 2-4i has the coordinates (2,-4)
-Point 6+i has the coordinates (6,1)
#We must find the distance between the two coordinates (2,-4) and (6,1):
Hence, the distance between the two points is 7.211
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
Answer:
A
Step-by-step explanation:
I'm not sure I understand the question. Is this worded correctly?
