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const2013 [10]
3 years ago
14

The table shows y as a function of x. Suppose a point is added to this table. Which choice gives a point that preserves the func

tion?
A) (−5, 7)
B) (9, −5)
C) (−1, −5)
D) (−8, −6)

Mathematics
1 answer:
Hunter-Best [27]3 years ago
8 0

Answer:

<em>A) (-5,7)</em>

Step-by-step explanation:

<u>Functions and Relations</u>

A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).

By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option

\boxed{A)\ (-5,7)}

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<h2>x = 3, y = 8</h2>

Step-by-step explanation:

We know:

The diagonals of the parallelogram are divided in half.

Therefore

EK ≅ KG and HK ≅ KF.

We have

EK = 7x + 5, KG = 3x + 17, HK = 3y - 11 and KF = y + 5

Equations:

(1)   7x + 5 = 3x + 17

(2) 3y - 11 = y + 5

Solutions:

(1)

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7x + 5 - 5 = 3x + 17 - 5

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7x - 3x = 3x - 3x + 12

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4x/4 = 12/4

x = 3

------------------------------

(2)

3y - 11 = y + 5                <em>add 11 to both sides</em>

3y - 11 + 11 = y + 5 + 11

3y = y + 16         <em>subtract y from both sides</em>

3y - y = y - y + 16

2y = 16           <em>divide both sides by 2</em>

2y/2 = 16/2

y = 8

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