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chubhunter [2.5K]
3 years ago
9

Is 6.7 mi more precise than 6 mi?

Mathematics
2 answers:
Nataly [62]3 years ago
8 0
The answer is . . . <u><em>no</em></u> (and the reason is because you haven't provided enough information data for one to determine the precision of the distance).

"6.7 mi" you might want to argue is more precise simply because it measures this distance to the tenth of a mile

"6 mi" happens to equal "6.0 mi" which is also measured to the tenth of a mile.

Precision of a value is determined on the basis of repeatability and/or reproducibility.  If the distance is measure 3 times with results that closely match each other - that's precision.

Unfortunately, you didn't provide us with enough information, so there's no way anyone could determine whether or not this measurement is precise.

Arisa [49]3 years ago
7 0
Yes because precise means specific.
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Answer:

A. (-1, 3)

Step-by-step explanation:

To solve, find which point in the options are in the double shaded region (the region that is shaded the darkest which is the region above the point where the lines cross).

A. (-1, 3) is the correct answer because it falls in the correct region.

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6 0
2 years ago
Use the identity (x^2+y^2)^2=(x^2−y^2)^2+(2xy)^2 to determine the sum of the squares of two numbers if the difference of the squ
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Answer:

Step-by-step explanation:

(x^2+y^2)^2=(x^2)^2+2x^2y^2+(y^2)^2

Adding and substracting 2x^2y^2

We get

(x^2+y^2)^2=(x^2)^2+2x^2y^2+(y^2)^2 +2x^2y^2-2x^2y^2

And we know a^2-2ab+b^2=(a-b)^2

So we identify (x^2)^2 as a^2 ,(y^2)^2 as b^2 and -2x^2y^2 as - 2ab. So we can rewrite (x^2+y^2)^2=(x^2 - y^2)^2 + 2x^2y^2 + 2x^2y^2= (x^2 - y^2)^2+4x^2y^2= (x^2 - y^2)^2+2^2x^2y^2

Moreever we know (a·b·c)^2=a^2·b^2·c^2 than means 2^2x^2y^2=(2x·y)^2

And (x^2+y^2)^2=(x^2 - y^2)^2 + (2x·y)^2

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