3 years ago

Which ordered pair could be removed to make this relation a function?

Mathematics

2 answers:

devlian [24]3 years ago

8 0

Answer:

(4,-2)

Step-by-step explanation:

garik1379 [7]3 years ago

6 0

Answer:

C- (4,-2)

Step-by-step explanation:

right edg 2021

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4 years ago

Perform the indicated operations below and simplify ((a+b)/4) + ((2a-b)/5).

ArbitrLikvidat [17]

Answer:

$\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{13a}{20} +\frac{b}{20}$

Step-by-step explanation:

We have:

$\frac{(a+b)}{4} +\frac{(2a-b)}{5}$

We can use <em>common denominator</em>.

Observation:

If you have, $\frac{a}{b} +\frac{c}{d}=\frac{(a*d)+(c*b)}{b*d}$

Then,

$\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{5(a+b)+4(2a-b)}{4*5}$

Using <em>distributive property</em>:

Observation:

c(a+b)=ca+cb

$\frac{5(a+b)+4(2a-b)}{4*5}=\frac{5a+5b+8a-4b}{20}=\frac{(5a+8a)+(5b-4b)}{20}$

Finally,

$\frac{(5a+8a)+(5b-4b)}{20} =\frac{13a+b}{20}=\frac{13a}{20} +\frac{b}{20}$

The answer then is:

$\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{13a}{20} +\frac{b}{20}$

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