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just olya [345]
3 years ago
6

A figure is reflected over the y-axis. Which of the following are true? The pre-image and image are congruent. The y-coordinates

' signs change. The x-coordinates' signs change. The x-axis is the line of reflection.
Mathematics
1 answer:
serg [7]3 years ago
3 0
When a reflection occurs, the original graph and the reflected graph are mirror images of each other. So, when the graph is reflected about the y-axis, think of the y-axis as the mirror. When you fold the graph length-wise, you will see that both graphs would coincide. Thus, their y-coordinate would not change. It would only change to the opposite quadrant, thereby changing the signs of the x-coordinates. So, as a rule, during reflection, the coordinates (x,y) is changed to (-x,y).

Therefore, the answer is:<span>The x-coordinates' signs change..</span>
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dimulka [17.4K]

m<1=23

m<2=90

m<3=67

m<4=113

m<5=67

the square means it is 90° so 2 is 90°.

all of the angles combined equals 360 so.

360-67-90= 203

since there is a straight line splitting down the middle. m<4=180-67=113

so m<4=113

now you would go 360-67-113-90=90 so

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180-90-67=23.

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aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 2

\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 3

\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\

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We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity \sin^2(\theta)+\cos^2(\theta) = 1 in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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