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3 years ago

The line segment is to be reflected over the line shown. What would be the length of the reflected line segment?

Mathematics

2 answers:

Andrej [43]3 years ago

5 0

Answer:

Step-by-step explanation:

In geometry (or math in general), when you do a reflection, rotation, or translation the size or length of the lines don't change. Think about it this way: if I have a log and I bring it into the bathroom with me (don't ask why idk) and hold it up in from of the mirror, what will happen? It will look the same length (and people will think I'm crazy for bringing a log into the restroom with me... but I digress).

anyanavicka [17]3 years ago

5 0

Answer:

The answer is "A"

Step-by-step explanation:

A reflection across a line is a rigid transformation. Therefore the shape of the original object will not change. Because of this, you know that it would have the same length as the original line segment.

I hope this helps!!!!

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3 years ago

In the right triangle below, tanA = 0.45. What is the approximate length of AB?

garik1379 [7]

In the right triangle ABC wherein AB is the hypotenuse, BC is the opposite and CA is the adjacent tanA=0.45. The approximate length of AB which is the hypotenuse is 22, Opposite (BC) is 9 and Adjacent (CA) is 20. You need to use pythagorean formula in getting the length of AB.

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3 years ago

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

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3 years ago

Twice a number is equal to negative four. Which equation could be used to find the number? 2n = 4 2n = -4n 2n - 4 2n = -4

Pepsi [2]

Twice a number is equal to negative four :
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answer
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