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mariarad [96]
3 years ago
8

How many different ways can the line be named? What are those names?

Mathematics
2 answers:
boyakko [2]3 years ago
5 0

Answer: A. 3 ways: k, DE, ED (both DE and ED have line markers over top)

To name a line, we just need two points on the line. We list them in any order because the line extends forever in both directions. Contrast this with a ray where order does matter. The little k is another way to name a line, potentially simplifying things.

Choice B is close, but it mentions ray DE instead of line DE. Choice C is missing line ED. Choice D is a similar story as choice B. These facts allow us to rule out B through D.

Genrish500 [490]3 years ago
3 0
A should be your answer
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In a fish tank , 8/11 of the fish have a red stripe on them. If 16 of the fish have red stripes, how man total fish are in the t
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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the
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Answer:

The wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} East of North  with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the  starting point in the direction of 35° E of South as shown in the figure.

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From figure, the magnitude of the required displacement vector is

|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)

and the direction is \alpha east of north as shown in the figure,

\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)

From the figure,

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\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|

\Rightarrow |AB|=45.52 miles

Again, |PQ|=|OP-OQ|

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\tan \alpha=\frac{|46.42|}{|45.52|}

\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}

Hence, the wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} E astof North  with respect to the destination point.

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