Line g is dilated by a scale factor of one half from the origin to create line g'. Where are points E' and F' located after dila
tion, and how are lines g and g' related?
A. The locations of E' and F' are E' (−8, 0) and F' (0, 4), and lines g and g' intersect at point F.
B. The locations of E' and F' are E' (−4, 0) and F' (0, 2), and lines g and g' are the same line.
C. The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.
D. The locations of E' and F' are E' (−1, 0) and F' (0, 0), and lines g and g' are not related.
A. The locations of E' and F' are E' (−8, 0) and F' (0, 4), and lines g and g' intersect at point F.
B. The locations of E' and F' are E' (−4, 0) and F' (0, 2), and lines g and g' are the same line.
C. The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.
D. The locations of E' and F' are E' (−1, 0) and F' (0, 0), and lines g and g' are not related.
2 answers:
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Let's start by visualising this concept.
Number of grains on square:
1 2 4 8 16 ...
We can see that it starts to form a geometric sequence, with the common ratio being 2.
For the first question, we simply want the fifteenth term, so we just use the nth term geometric form:
Thus, there are 16, 384 grains on the fifteenth square.
The second question begs the same process, only this time, it's a summation. Using our sum to n terms of geometric sequence, we get:
Thus, there are 32, 767 total grains on the first 15 squares, and you should be able to work the rest from here.
Number of grains on square:
1 2 4 8 16 ...
We can see that it starts to form a geometric sequence, with the common ratio being 2.
For the first question, we simply want the fifteenth term, so we just use the nth term geometric form:
Thus, there are 16, 384 grains on the fifteenth square.
The second question begs the same process, only this time, it's a summation. Using our sum to n terms of geometric sequence, we get:
Thus, there are 32, 767 total grains on the first 15 squares, and you should be able to work the rest from here.