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Neporo4naja [7]
3 years ago
13

Which of these points does not change its location when it is reflected across the y-axis? A (2, 0) b (0, 6) c (3, 3) d (5, 5)

Mathematics
2 answers:
yanalaym [24]3 years ago
5 0

B because it has the point 0 which can not be negative when reflected across the y axis like 2, 3, and 5 can.
Andru [333]3 years ago
4 0

Answer:

B. (0,6)

Step-by-step explanation:

To solve this question, you'll need to figure out what the final point is after reflecting each point. A quick way to figure this out is by multiplying (-1) to the x-coordinate.

When you reflect A. (2,0) over the y-axis, the point becomes (-2,0).

When you reflect C. (3,3) over the y-axis, the point becomes (-3,3).

When you reflect D. (5,5) over the y-axis, the point becomes (-5,5).

The only answer that does not change location is B. (0,6) as it stays at (0,6). If you multiply 0 by any number, it will always stay 0.

Multiplying the x-coordinate by (-1) to find the reflection point only works if you are reflecting it over the y-axis. You would multiple the y-coordinate by (-1) if you were reflecting over the x-axis.

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Answer:

<h2>The distance from the pitcher's mound and to second base is 37.99 approximately.</h2>

Step-by-step explanation:

The diamond is a square, which in this case has 50 feet long each side, and from home to pitcher is 38 feet. Notice that home is a vertex of the square and the pitcher's mound is the intersection of the diagonals, where they cut half.

We can find the distance from the pitcher to first base using Pythagorean's Theorem, where 50 feet is the hypothenuse.

50^{2} =38^{2}+x^{2}\\x^{2}=50^{2}-38^{2}\\x=\sqrt{2500-1444}\\ x=\sqrt{1056}\\ x \approx 32.5 \ ft

Therefore, the distance from the pitcher to first base is 32.5 feet, approximately.

Now, we can use again Pythagorean's Theorem to find the distance from pitcher to second base, where the hypothenuse is 50 feet.

50^{2}=32.5^{2}+y^{2}\\y^{2}=50^{2}-32.5^{2}\\y=\sqrt{2500-1056.25}\\ y =\sqrt{1443} \approx 37.99

Therefore, the distance from the pitcher's mound and to second base is 37.99 approximately.

<em>(this results make sense, because the diagonals of a square intersect at half, that means all bases have the same distance from pitcher's mound, so the second way to find the distance asked in the question is just using theory)</em>

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