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4 years ago

Hexagon ABCDEF → hexagon A'B'C'D'E'F'. How can the transformation shown be described? Check all that apply.

Mathematics

2 answers:

netineya [11]4 years ago

7 0

Answer:

I agree that it is C and F

Step-by-step explanation:

AveGali [126]4 years ago

5 0

The correct answer would be c,and f

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Step-by-step explanation:

-5-(3^2+2)

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Step-by-step explanation:

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3 years ago

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zhenek [66]

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<u>Step-by-step explanation:</u>

Inverse is when you swap the x's and y's and solve for y:

$y+\dfrac{13}{4}=\bigg(x-\dfrac{1}{2}\bigg)^2\\\\\\x+\dfrac{13}{4}=\bigg(y-\dfrac{1}{2}\bigg)^2\rightarrow \text{(swapped the x's and y's)}\\\\\\\sqrt{x+\dfrac{13}{4}}=\sqrt{\bigg(y-\dfrac{1}{2}\bigg)^2}\rightarrow \text{(took square root of both sides)}\\\\\\\sqrt{\dfrac{4x+13}{4}}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(created common denominator in radical)}\\\\\\ \dfrac{\sqrt{4x+13}}{2}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(simplified radical)}$

$\pm \dfrac{\sqrt{4x+13}}{2}=y-\dfrac{1}{2}}\rightarrow \text{(divided both sides by}\ \pm )\\\\\\\dfrac{1}{2}\pm \dfrac{\sqrt{4x+13}}{2}=y\quad \rightarrow \text{(added}\ \dfrac{1}{2}\ \text{to both sides)}\\\\\\\dfrac{1\pm \sqrt{4x+13}}{2}=y\quad \rightarrow \text{(combined numerators)}$

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3 years ago

Which set of ordered pairs represents a function?

soldi70 [24.7K]

Answer:

(1,2) and (1,5)

Step-by-step explanation:

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4 years ago