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BlackZzzverrR [31]
1 year ago
10

What set of reflections would carry triangle ABC onto itself?

Mathematics
2 answers:
aivan3 [116]1 year ago
8 0
<span><u><em>The correct answer is:</em></u>
4) y-axis, x-axis, y-axis, x-axis.

<u><em>Explanation</em></u><span><u><em>: </em></u>
Reflecting a point (x,y) across the <u>x-axis</u> will map it to (x,-y).
Reflecting a point (x,y) across the <u>y-axis</u> will map it to (-x,y).
Reflecting a point (x,y) across the line <u>y=x</u> will map it to (y, x).

We want a series of transformations that will map every point (x,y) back to (x,y). This means that everything that gets done in one transformation must be undone in another. The only one where this happens is #4.

Reflecting across the y-axis first negates the x-coordinate; (x,y) goes to (-x,y).
Reflecting this across the x-axis negates the y-coordinate; (-x,y) goes to (-x,-y).
Reflecting this point back across the y-axis negates the x-coordinate again, returning it to the original: (-x,-y) goes to (x,-y).
Reflecting this point back across the x-axis negates the y-coordinate again, returning it to the original: (x,-y) goes to (x,y).
We are back to our original point.</span></span>
8_murik_8 [283]1 year ago
3 0

Answer: 4) y-axis, x-axis, y-axis, x-axis

Step-by-step explanation:

Let (x,y) represents the coordinates of the triangle ABC,

1) Under the set of reflection, x-axis, y=x, y-axis, x-axis,

(x,y)\rightarrow (x,-y)\rightarrow (-x,-(-y))\rightarrow (-(-x),y)\rightarrow (x,-y)

Hence, (x,-y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections does not carry triangle ABC onto itself.

2) Under the set of reflection,  x-axis, y-axis, x-axis

(x,y)\rightarrow (x,-y)\rightarrow (-x,-y))\rightarrow (-x,y)

Hence, (-x,y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections does not carry triangle ABC onto itself.

3) Under the set of reflection, y=x, x-axis, x-axis

(x,y)\rightarrow (-x,-y)\rightarrow (-x,-(-y))\rightarrow (-x,-y)

Hence, (-x,-y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections does not carry triangle ABC onto itself.

4) Under the set of reflection, y-axis, x-axis, y-axis, x-axis

(x,y)\rightarrow (-x,y)\rightarrow (-x,-y)\rightarrow (-(-x),-y)\rightarrow (x,-(-y))

Hence, (x,y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections carries triangle ABC onto itself.

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earnstyle [38]

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The answer is 3√5 mi.

The formula is: d = √(3h/2)

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Shawn:

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How much farther can Shawn see to the horizon?

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3 0
2 years ago
If LO = 15x+19 and QN = 10x+2 find PN
svet-max [94.6K]

Answer:

PN=64\ units

Step-by-step explanation:

<u><em>The complete question is</em></u>

Given the quadrilateral is a rectangle, if LO = 15x+19 and QN = 10x+2 find PN

see the attached figure to better understand the problem

we know that

The diagonals of a rectangle are congruent and bisect each other

so

QN=\frac{1}{2}LO

substitute the given values

10x+2=\frac{1}{2}(15x+19)

solve for x

20x+4=15x+19\\20x-15x=19-4\\5x=15\\x=3

Find the length of PN

Remember that

PN=LO ----> diagonals of rectangle are congruent

LO=15x+19

substitute the value of x

LO=15(3)+19=64\ units

therefore

PN=64\ units

8 0
2 years ago
Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu
NeX [460]

Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

Let speed in clear weather = x

⇒ Speed in thunderstorm = x-20

Total time taken for trip = 1.5 hours

We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm

⇒ Total time taken for trip = \frac{Distance covered in clear weather}{Speed in clear weather} + \frac{Distance covered in thunderstorm}{Speed in thunderstorm}

⇒ 1.5 = \frac{50}{x} + \frac{15}{x-20}

⇒ 1.5 = \frac{50(x-20)+15(x)}{(x)(x-20)}

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = \frac{-b+\sqrt{D}}{2a} ; x₂ = \frac{-b-\sqrt{D}}{2a}

⇒ x₁ = \frac{-(-190)+\sqrt{12,100}}{2*3} and x₂ = \frac{-(-190)-\sqrt{12,100}}{2*3}

⇒ x₁ = \frac{190+110}{6} and x₂ = \frac{190-110}{6}

⇒ x₁ = \frac{300}{6} and x₂ = \frac{80}{6}

⇒ x₁ = 50 and x₂ = 13.33

So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.

If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

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