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Andru [333]
3 years ago
11

Which point could be removed in order to make the relation a function? (–9, –8), (–{(8, 4), (0, –2), (4, 8), (0, 8), (1, 2)}

Mathematics
2 answers:
stepan [7]3 years ago
7 0

We are given order pairs  (–9, –8), (–{(8, 4), (0, –2), (4, 8), (0, 8), (1, 2)}.

We need to remove in order to make the relation a function.

<em>Note: A relation is a function only if there is no any duplicate value of x coordinate for different values of y's of the given relation.</em>

In the given order pairs, we can see that (0, –2) and (0, 8) order pairs has same x-coordinate 0.

<h3>So, we need to remove any one (0, –2) or (0, 8) to make the relation a function.</h3>
Alex17521 [72]3 years ago
3 0
I'm gonna say (0,8) because there can't be two of the same x values in a function.
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larisa86 [58]

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Step-by-step explanation:

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3 years ago
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salantis [7]

Check the first picture below.


for a parallelogram, it has to have two pairs of parallel sides, so BC || AD and AB || CD.

if two lines are parallel, they have the same slope, namely the <u>same rise and run</u>.

notice, the line BC has a run of 6 and a rise of 1, <u>arrows in red</u>, therefore, AD must also have a run of 6 and a rise of 1, so if we move from A 6 units over and 1 up, we'd end up at point D.


now, for part atop

Check the 2nd picture, in a parallelogram, opposite sides are parallel and opposite angles are equal, so if WLP is 144°, then PNW is also 144°.

in a parallelogram, diagonals bisect each other, so each cut the other in halves, so if PM is 9 then MW is also 9 and thus PW is just 9+9=18.


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notice what we said about a parallelogram above, thus


\bf 8x+13=11x-23\implies 8x+36=11x\implies 36=3x \\\\\\ \cfrac{36}{3}=x\implies \boxed{12=x} \\\\[-0.35em] ~\dotfill\\\\ 2y+12=4y-4\implies 2y+16=4y\implies 16=2y \\\\\\ \cfrac{16}{2}=y\implies \boxed{8=y}


so then TQ ==> 2(8) + 12 ==> 16 + 12 ==> 28.

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and since in a parallelogram, adjacent interior angles add up to 180°, then ∡Q + ∡T = 180°.

∡T ==> 180 - 109 ==> 71°.

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