How do you solve this?
![How do you solve this class=](https://us-static.z-dn.net/files/d6a/8a86970bfe02fa1c82cf639e64ae98bd.png)
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Explanation:
If we plugged (x,y) = (0,8) into the first inequality, then we get
y < x^2+6
8 < 0^2+6
8 < 0+6
8 < 6
which is false. So we can rule out choice A.
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Trying choice B leads us to
y < x^2+6
2 < 4^2+6
2 < 16+6
2 < 22
That last statement is true, so the first inequality is true for (x,y) = (4,2)
Let's try the other inequality
y > x^2-4
2 > 4^2-4
2 > 16-4
2 > 12
That's false. Since one of the inequalities is false (it doesn't matter which one), this means the entire system is false. We cross choice B off the list.
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Now onto choice C
You should find that y < x^2+6 becomes -4 < 10 after plugging in (x,y) = (-2,-4). Since -4 < 10 is true, we move onto the next inequality.
The inequality y > x^2-4 becomes -4 > 0 after plugging in those mentioned x,y values. The inequality -4 > 0 is false.
We cross choice C off the list.
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The only thing left is choice D. It has to be the answer.
Let's find out if we get true inequalities when plugging in (x,y) = (2,6)
y < x^2+6
6 < 2^2+6
6 < 10 ... true
and
y > x^2-4
6 > 2^2-4
6 > 4-4
6 > 0 .... also true
Both inequalities are true, so the entire system is true. Therefore, (x,y) = (2,6) is one of the infinitely many solutions to this system.
Choice D is confirmed as the answer
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Refer to the diagram below. I've graphed the two dashed boundary curves and the shaded region between. This region is above the y = x^2-4 curve, and below the y = x^2+6 curve. So we're ignoring the stuff above the y = x^2+6 curve.
Points A through D represent the four answer choices in the order given. We see that point D is the only point in the shaded region, so that visually confirms we have the correct answer.
Note: points on the dashed boundaries do not count as solutions.