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Aleonysh [2.5K]
3 years ago
12

What set of reflections would carry rectangle ABCD onto itself?

Mathematics
1 answer:
Furkat [3]3 years ago
7 0
Answer: option <span>D) y=x, x-axis, y=x, y-axis</span>.

I first thought it was the option C) and I tried with it but it was wrong. This is how I dit it.

Option C step by step:

<span>1) Reflection over the x - axis => point with coordinates (a,b) is transformed into point with coordinates (a, -b)

2) Reflection over the line y = x  => point with coordinates (a, -b) is transformed into point with coordinates (-b,a)

3) New feflection over the x - axis => (-b,a) transforms into (-b, -a)

4) New reflection over the line y = x => (-b,-a) transforms into (-a,-b)

Which shows it is not the option C).

Then I probed with option D. Step by step:

1) Reflection over the line y = x => (a,b) → (b,a)

2) Reflection over the x-axis => (b,a) → (b,-a)

3) Reflection over the line y = x => (b,-a) → (-a,b)

4) Reflection over the y-axis => (-a,b) → (a,b).

So, this set of reflections, given by the option D) transforms any point into itself, which proofs that the option D) is the right answer.
</span>
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I thought of a number, I added 1 3 4 to it, then I multiplied the result by 2 2 11 and I got 8. What was my number?
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Given:

In a number, adding 1\dfrac{3}{4}, then multiplying the result by 2\dfrac{2}{11} we get 8.

To find:

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Solution:

Let the unknown number be x.

According to the question,

(x+1\dfrac{3}{4})\times 2\dfrac{2}{11}=8

(x+\dfrac{1\times 4+3}{4})\times \dfrac{2\times 11+2}{11}=8

(x+\dfrac{4+3}{4})\times \dfrac{22+2}{11}=8

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Multiply both sides by \dfrac{11}{24}.

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Subtract \dfrac{7}{4} from both sides.

x=\dfrac{11}{3}-\dfrac{7}{4}

x=\dfrac{44-21}{12}

x=\dfrac{23}{12}

x=1\dfrac{11}{12}

Therefore, the required number is 1\dfrac{11}{12}.

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