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Keith_Richards [23]
3 years ago
12

Point (2, -4) is reflected across the x-axis. What are the coordinates of its image?

Mathematics
2 answers:
Zina [86]3 years ago
7 0

The coordinates of the image of point (2 , -4) after reflection across

the x-axis is (2 , 4)

Step-by-step explanation:

Let us revise the reflection of a point

1. If point (x , y) reflected across the x-axis, then its image is (x , -y)

2. If point (x , y) reflected across the y-axis, then its image is (-x , y)

3. If point (x , y) reflected across the line y = x, then its image is (y , x)

4. If point (x , y) reflected across the line y = -x, then its image is (-y , -x)

∵ Point (2, -4) is reflected across the x-axis

∵ Point (x , y) reflected across the x-axis, then its image is (x , -y)

- Chang the sign of the y-coordinate of the point to find its image

∵ The y-coordinate of the point is -4

∴ The y-coordinate of the image is 4

∴ The image of the point (2 , -4) is (2 , 4) after reflection across the

   x-axis

The coordinates of the image of point (2 , -4) after reflection across

the x-axis is (2 , 4)

Learn more:

You can learn more about reflection in brainly.com/question/5017530

#LearnwithBrainly

derrickw22 years ago
0 0

Using what you've learned, what would the image of (2, -4) be if reflected over the y-axis?

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Divide the first equation by 5

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Subtract the first equation from the second

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x + 2y = 3

Subtract the first equation from the second again

x        = 1

   2y = 2

Divide the second equation by 2

x        = 1

      y = 1

<h3>So, the solution is  x = 1  and  y = 1  {or: (1, 1)} </h3>
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Salem High School's winter sports teams WON 72% of their games last season. If the teams played 50 games, how many games did the
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Answer

Find out the how many games did the teams lose .

To prove

Formula

Percentage = \frac{Part\ value\times 100}{Total\ values}

As given

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Number of  games teams loses = Total number of games - number of games win

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{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

‣ A coin is tossed three times.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

‣ The probability of getting,

1) Exactly 3 tails

2) At most 2 heads

3) At least 2 tails

4) Exactly 2 heads

5) Exactly 3 heads

{\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}

\star \: \tt  P(E)= {\underline{\boxed{\sf{\red{  \dfrac{ Favourable \:  outcomes }{Total \:  outcomes}  }}}}}

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

★ When three coins are tossed,

then the sample space = {HHH, HHT, THH, TTH, HTH, HTT, THT, TTT}

[here H denotes head and T denotes tail]

⇒Total number of outcomes \tt [ \: n(s) \: ] = 8

<u>1) Exactly 3 tails </u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly  \: 3 \:  tails)}  =  \red{ \dfrac{1}{8}}

<u>2) At most 2 heads</u>

[It means there can be two or one or no heads]

Here

• Favourable outcomes = {HHT, THH, HTH, TTH, HTT, THT, TTT} = 7

• Total outcomes = 8

\therefore  \sf Probability_{(at \: most  \: 2 \:  heads)}  =  \green{ \dfrac{7}{8}}

<u>3) At least 2 tails </u>

[It means there can be two or more tails]

Here

• Favourable outcomes = {TTH, TTT, HTT, THT} = 4

• Total outcomes = 8

\longrightarrow   \sf Probability_{(at \: least \: 2 \:  tails)}  =  \dfrac{4}{8}

\therefore  \sf Probability_{(at \: least \: 2 \:  tails)}  =   \orange{\dfrac{1}{2}}

<u>4) Exactly 2 heads </u>

Here

• Favourable outcomes = {HTH, THH, HHT } = 3

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 2 \:  heads)}  =  \pink{ \dfrac{3}{8}}

<u>5) Exactly 3 heads</u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 3 \:  heads)}  =  \purple{ \dfrac{1}{8}}

\rule{280pt}{2pt}

8 0
2 years ago
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