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Burka [1]
3 years ago
14

Two transformations are performed on a figure on a coordinate plane. The first transformation is translated 6 units up and 3 uni

ts right. Which second transformation will result in an image that is similar to, but not congruent to, the original figure?

Mathematics
2 answers:
Bess [88]3 years ago
7 0

Answer:

The last answer choice

Step-by-step explanation:

This transformation doesn't not keep its congruent

Marina CMI [18]3 years ago
4 0
3nd answer is the right answer for that question.
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3 0
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To show me similarity to this statement, how can it be done?
Alenkasestr [34]

We start with the expression at the left of the equation.

We can combine the terms as:

\begin{gathered} \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}} \\ \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})} \end{gathered}

We can now apply the distributive property for the both the numerator and denominator. We can see also that the denominator is the expansion of the difference of squares:

\begin{gathered} \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2})^2-(\sqrt[]{2-\sqrt[]{3}}))^2} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})+(\sqrt[]{3}-2)\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{2^{}-(2-\sqrt[]{3})^{}} \\ \frac{\sqrt[]{2}\cdot(2+\sqrt[]{3})-\sqrt[]{2-\sqrt[]{3}}\cdot(2+\sqrt[]{3})+\sqrt[]{2}\cdot(\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}\cdot(\sqrt[]{3}-2)}{2-2+\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2+\sqrt[]{3}+\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}(-2-\sqrt[]{3}+\sqrt[]{3}-2)}{\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2\sqrt[]{3})+\sqrt[]{2-\sqrt[]{3}}(-4)}{\sqrt[]{3}} \\ 2\sqrt[]{2}-4\frac{\sqrt[]{2-\sqrt[]{3}}}{\sqrt[]{3}} \end{gathered}

We then can continue rearranging this as:

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1 year ago
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