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3 years ago

Which sequences of transformations confirm the congruence of shape II and shape I?

Mathematics

2 answers:

Pani-rosa [81]3 years ago

7 0

Answer: the answer is B.

mylen [45]3 years ago

6 0

Answer:

1. When we reflect the shape I along X axis it will take the shape I in first quadrant, and then if we rotate the shape I by 90° clockwise, it will take the shape again in second quadrant . So we are not getting shape II. This Option is Incorrect.

2. Second Option is correct , because by reflecting the shape I across X axis and then by 90° counterclockwise rotation will take the Shape I in second quadrant ,where we are getting shape II.

3. a reflection of shape I across the y-axis followed by a 90° counterclockwise rotation about the origin takes the shape I in fourth Quadrant. →→ Incorrect option.

4. This option is correct, because after reflecting the shape through Y axis ,and then rotating the shape through an angle of 90° in clockwise direction takes it in second quadrant.

5. A reflection of shape I across the x-axis followed by a 180° rotation about the origin takes the shape I in third quadrant.→→Incorrect option

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Hii! I need some help, would anyone mind helping me?

Kisachek [45]

Step-by-step explanation:

${x}^{2} - 14x - 25 {y}^{2} + 100y = 76$

Factor by grouping,

$( {x}^{2} - 14x) - (25 {y}^{2} - 100y) = 76$

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$( {x}^{2} - 14x + 49) - (25 {y}^{2} - 100y) = 125$

Factor out 25 for the y variables

$( {x}^{2} - 14x + 49) - 25( {y}^{2} - 4y) = 125$

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$( {x}^{2} - 14x + 49) - 25( {y}^{2} - 4y + 4) = 25$

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$(x - 7) {}^{2} - 25(y - 2) {}^{2} = 25$

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2 years ago

PLEASE HELP ASAP IT IS VERY MUCH APPRECIATED

irakobra [83]

YOUR CORRECT good luckkk

5 0

3 years ago

Read 2 more answers

what is the polynomial function of lowest degree with leading coefficient of 1 and roots 2 and 1 + square root2

Marta_Voda [28]

If the roots to such a polynomial are 2 and $1+\sqrt2$ , then we can write it as

$(x-2)(x-(1+\sqrt2))$

courtesy of the fundamental theorem of algebra. Now expanding yields

$(x-2)(x-1-\sqrt2)=x^2-2x-(1+\sqrt2)x+2(1+\sqrt2)=x^2-(3+\sqrt2)x+2+2\sqrt2$

which would be the correct answer, but clearly this option is not listed. Which is silly, because none of the offered solutions are *the* polynomial of lowest degree and leading coefficient 1.

So this makes me think you're expected to increase the multiplicity of one of the given roots, or you're expected to pull another root out of thin air. Judging by the choices, I think it's the latter, and that you're somehow supposed to know to use $1-\sqrt2$ as a root. In this case, that would make our polynomial

$(x-2)(x-(1+\sqrt2))(x-(1-\sqrt2))=x^3-4x^2+3x+2$

so that the answer is (probably) the third choice.

Whoever originally wrote this question should reevaluate their word choice...

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