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Roman55 [17]
2 years ago
5

∆ABC transforms to produce ∆A'B'C'. Which transformation did NOT take place?

Mathematics
2 answers:
never [62]2 years ago
6 0

Answer: The answer is (D) Reflection across the line y = -x.


Step-by-step explanation:  In figure given in the question, we can see two triangles, ΔABC and ΔA'B'C' where the second triangle is the result of transformation from the first one.

(A) If we rotate ΔABC 180° counterclockwise about the origin, then the image will coincide with ΔA'B'C'. So, this transformation can take place here.

(B) If we reflect ΔABC across the origin, then also the image will coincide with ΔA'B'C' and so this transformation can also take place.

(C) If we rotate ΔABC through 180° clockwise about the origin, the we will see the image will be same as ΔA'B'C'. Hence, this transformation can also take place.

(D) Finally, if we reflect ΔABC across the line y = -x, the the image formed will be different from ΔA'B'C', in fact, it is ΔA'D'E', as shown in the attached figure. So, this transformation can not take place here.

Thus, the correct option is (D).


Black_prince [1.1K]2 years ago
4 0

Answer:

D.reflection across the line y = -x

Step-by-step explanation:

I had the same question asked on a assignment

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Answer:

x = 20

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Step-by-step explanation:

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P = -0.5x² + 80x - 700

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This is in the standard form 0 = ax² + bx + c

Use the quadratic formula to find values of x

x = \frac{-b±\sqrt{b^{2}-4ac}  }{2a}

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a = -0.5

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x = \frac{-(80)±\sqrt{80^{2}-4(-0.5)(-1400)}  }{2(-0.5)}

x = \frac{-80±60}{-1}

Split the formula at the ± so that there are two to get the two x values.

x = \frac{-80+60}{-1}

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The profit will be $700 when x is 20 or when x is 140.

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\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_1^2(-u\cos v\,\vec\imath-u\sin v\,\vec\jmath+u^3\,\vec k)\cdot(u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k)\,\mathrm du\,\mathrm dv

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Hello,

a) | |PA|-π |>| |PB|-π | (see pic)

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