Which pairs of quadrilaterals can be shown to be congruent using rigid motions?
2 answers:
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<h3>Answer: n = -11</h3>
The other choices are false, so choice C is the only solution here.
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Explanation:
Draw a number line. You'll have 0 in the center, with positive values on the right. The positive values increase 1,2,3,4,... as you move from left to right.
The negative values will go in the opposite direction. When going from right to left, we have -1, -2, -3, -4, ...
See the diagram below.
Now plot an open hole at -7 on the number line. The open hole says that we exclude this value from the solution set. Then we shade everything to the left of the open circle. This visually describes all values smaller than -7. Something like -2 is not to the left of -7, meaning -2 < -7 is false. But a value like -10 is to the left of -7, making -10 < -7 true.
With that in mind, the answer is n = -11 since -11 is to the left of -7. Of the values in the answer choices, it's the only thing to the left of -7.
The value n = -7 cannot be the solution because n < -7 would become -7 < -7 which is false.