Quadrilateral EFGH was dilated by a scale factor of 2 from the center (1, 0) to create E'F'G'H'. Which characteristic of dilatio
ns compares segment F'H' to segment FH?
A). A segment that passes through the center of dilation in the pre-image continues to pass through the center of dilation in the image.
B). A segment that does not pass through the center of dilation in the pre-image is parallel to its corresponding segment in the image.
C). A segment in the image has the same length as its corresponding segment in the pre-image.
D). A segment that does not pass through the center of dilation in the pre-image is perpendicular to its corresponding segment in the image.
A). A segment that passes through the center of dilation in the pre-image continues to pass through the center of dilation in the image.
B). A segment that does not pass through the center of dilation in the pre-image is parallel to its corresponding segment in the image.
C). A segment in the image has the same length as its corresponding segment in the pre-image.
D). A segment that does not pass through the center of dilation in the pre-image is perpendicular to its corresponding segment in the image.
1 answer:
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Answer:
n = 8
Step-by-step explanation:
"By inspection" is an appropriate method.
We are asked to compare the expressions
n·n
8·n
and find the value(s) of n that makes them equal. <em>By inspection</em>, we see that n=8 will make these expressions equal. We also know that both expressions will be zero when n=0.
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More formally, we could write ...
n^2 = 8n . . . . the two formulas give the same value
n^2 -8n = 0 . . . . rearrange to standard form
n(n -8) = 0 . . . . . factor
Using the zero product rule, we know the solutions will be the values of n that make the factors zero. Those values are ...
n = 0 . . . . . makes the factor n = 0
n = 8 . . . . . makes the factor (n-8) = 0
Generally, we're not interested in "trivial" solutions (n=0), so the only value of n that is of interest is n = 8.
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A lot of times, I find a graphing calculator to be a quick and easy way to find function argument values that make expressions equal.
