<h3>
Answer: Choice D. (3j, 3k) and (3/j, 3/k)</h3>
The slash indicates a fraction.
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Proof:
We'll need to consider 4 different cases.
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Case (1): j > 0 and k > 0
If j > 0, then 3j > 0 and 3/j > 0
If k > 0, then 3k > 0 and 3/k > 0
The two points (3j, 3k) and (3/j, 3/k) are both in quadrant 1.
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Case (2): j > 0 and k < 0
If j > 0, then 3j > 0 and 3/j > 0
If k < 0, then 3k < 0 and 3/k < 0
Points (3j, 3k) and (3/j, 3/k) are both in quadrant 4.
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Case (3): j < 0 and k > 0
If j < 0, then 3j < 0 and 3/j < 0
If k > 0, then 3k > 0 and 3/k > 0
Points (3j, 3k) and (3/j, 3/k) are in quadrant 3.
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Case (4): j < 0 and k < 0
If j < 0, then 3j < 0 and 3/j < 0
If k < 0, then 3k < 0 and 3/k < 0
Points (3j, 3k) and (3/j, 3/k) are in quadrant 4.
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For nonzero integers j and k, we've shown that Points (3j, 3k) and (3/j, 3/k) are in the same quadrant. This concludes the proof.
![\frac{V_1}{V_2}=k^3 \\ \\ \frac{320 \ cm^3}{1080 \ cm^3}=k^3 \\ \frac{8}{27}=k^3 \\ k=\sqrt[3]{\frac{8}{27}} \\ \boxed{k=\frac{2}{3}} \Leftarrow \hbox{answer B}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BV_2%7D%3Dk%5E3%20%5C%5C%20%5C%5C%0A%5Cfrac%7B320%20%5C%20cm%5E3%7D%7B1080%20%5C%20cm%5E3%7D%3Dk%5E3%20%5C%5C%0A%5Cfrac%7B8%7D%7B27%7D%3Dk%5E3%20%5C%5C%0Ak%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B8%7D%7B27%7D%7D%20%5C%5C%0A%5Cboxed%7Bk%3D%5Cfrac%7B2%7D%7B3%7D%7D%20%5CLeftarrow%20%5Chbox%7Banswer%20B%7D)