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Natalija [7]
3 years ago
10

3. Does it matter which number comes first in an ordered pair? Explain.​

Mathematics
2 answers:
satela [25.4K]3 years ago
6 0

Answer:

yes

Step-by-step explanation:

This is because if you have the order pair (5,2) and you go up first instead of over first then that would be the order pair (2,5) wich would be a different point on the graph. Always go over and then up.

Alchen [17]3 years ago
3 0

Yes, the order of the numbers in an ordered pair does matter and here’s why:

If you think about an ordered pair, you know that the first number is the x-coordinate and the second number is the y-coordinate. On a graph, the x-coordinate tells you how far left or right the point is (positive goes right and negative goes left) and the y-coordinate tells you how far up or down the point is (positive goes up and negative goes down). If you have an ordered pair (6,2) for example, if you started at the origin (0,0) of a coordinate plane, you would go 6 units to the right and 2 units up. If you flipped the numbers (2,6) you would the go 2 units to the right and 6 units up. These two points (6,2) and (2,6) are not the same, so the order does matter.

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3 years ago
Find [-4/5]×3/7×15/16×[-14/9]​
dangina [55]

The given expression is \frac{-4}{5} \times \frac{3}{7}  \times \frac{15}{16}  \times (\frac{-14}{9}).

We need to simplify the given expression.

<h3>What is a rational number?</h3>

In mathematics, a rational number is a number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q.

Now group the rational numbers (\frac{-4}{5}\times \frac{15}{16} ) \times (\frac{3}{7}\times (\frac{-14}{9}))

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=(\frac{-1}{4}\times \frac{3}{1} ) \times (\frac{1}{3}\times (\frac{-2}{1}))

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Therefore, the simplified value of the given expression is \frac{1}{2}.

To learn more about the rational numbers visit:

brainly.com/question/17450097.

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8 0
2 years ago
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Free_Kalibri [48]

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8 and 10 can both be divided by 2. So, 8 ÷ 2 is 4 and 10 ÷ 2 is 5.

Hope this helps :)

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