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jolli1 [7]
3 years ago
8

DEF with vertices D(2, 5), E(6, 4), and F(3, 3), is reflected across the line y=x

Mathematics
1 answer:
snow_tiger [21]3 years ago
4 0
The student's answer is wrong. When a point (a, b) is reflected across the line y=x, the coordinates of the new point is reversed to (b, a). So D(2, 5) would become (5, 2), E(6, 4) would become (4, 6), and F(3, 3) is still (3, 3). None of the new points is (2, -5).
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Answer:

No

Step-by-step explanation:

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3 years ago
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Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =
Rainbow [258]

Answer:

\sin L = 0.60

tan\ N = 1.33

\cos L = 0.80

\sin N = 0.80

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

MN = 6

LN = 10

First, we need to calculate length LM,

Using Pythagoras theorem:

LN^2 = MN^2 + LM^2

10^2 = 6^2 + LM^2

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Collect Like Terms

LM^2 = 100 - 36

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Solving (a): \sin L

\sin L = \frac{Opposite}{Hypotenuse}

\sin L = \frac{MN}{LN}

Substitute values for MN and LN

\sin L = \frac{6}{10}

\sin L = 0.60

Solving (b): tan\ N

tan\ N = \frac{Opposite}{Adjacent}

tan\ N = \frac{LM}{MN}

Substitute values for LM and MN

tan\ N = \frac{8}{6}

tan\ N = 1.33

Solving (c): \cos L

\cos L = \frac{Adjacent}{Hypotenuse}

\cos L = \frac{LM}{LN}

Substitute values for LN and LM

\cos L = \frac{8}{10}

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Solving (d): \sin N

\sin N = \frac{Opposite}{Hypotenuse}

\sin N = \frac{LM}{LN}

Substitute values for LM and LN

\sin N = \frac{8}{10}

\sin N = 0.80

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Answer:

Step-by-step explanation:

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oksian1 [2.3K]

Answer:

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Step-by-step explanation:

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