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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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Step-by-step explanation:

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Recall the angle sum identities:

\sin(x+y)=\sin x\cos y+\cos x\sin y

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}

Divide through numerator and denominator by \cos x\cos y to get

\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}

Next, we use the fact that x,y lie in the first quadrant to determine that

\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2

\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}

So we then have

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1

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