Graph the image of the given triangle after the transformation with the rule (x, y)→(x, −y) .

Mathematics

2 answers:

Zarrin [17]3 years ago

4 0

Firstly, we will select three corner points

A=(2,3)

B=(6,8)

C=(7,4)

we are given

the transformation with the rule (x, y)→(x, −y)

y--->-y

so, it is reflected about x-axis

so, we will multiply y-value by -1

we get new points as

A=(2,-3)

B=(6,-8)

C=(7,-4)

now, we can locate these points and draw graph

we get

Lisa [10]3 years ago

4 0

Answer:

I just took the test and the answer is shown in the picture below!

Also thank you to the person above for getting the right answer!

You might be interested in

On a unit circle, the vertical distance from the x-axis to a point on the perimeter of the circle is twice the horizontal distan

swat32

Check the picture below.

since the vertical distance, namely the y-coordinate, is twice as much as the horizontal, then if the horizontal is "x", the vertical one must be 2x.

let's find the hypotenuse first.

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}$