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3 years ago

Point A is located at (3, -5). After it is transformed, point A' is located at (3, 5). How was the point transformed?

Mathematics

1 answer:

ASHA 777 [7]3 years ago

3 0

Answer:

Step-by-step explanation:

Under a reflection in the x- axis

a point (x, y ) → (x, - y )

Under a rotation about the origin of 90°

a point (x, y ) → (- y, x )

Under a reflection in the y- axis

a point (x, y ) → (- x, y )

Under a rotation of 180°

a point (x, y ) → (- x. - y )

Here A(3, - 5 ) → A'(3, 5 ) → A

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3 years ago

Find the equation of the inverse.<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%7B2%7D%5E%7Bx%20%2B%205%7D%20%20-%206" id="TexF

Contact [7]

Find the equation of the inverse.

$y = {2}^{x + 5} - 6$

we have

$y=\mleft\{2\mright\}^{\mleft\{x+5\mright\}}-6$

step 1

Exchange the variables

x for y and y for x

$x=\{2\}^{\{y+5\}}-6$

step 2

Isolate the variable y

$(x+6)=2^{(y+5)}$

apply log both sides

$\begin{gathered} \log (x+6)=(y+5)\cdot\log (2) \\ y+5=\frac{\log (x+6)}{\log (2)} \\ \\ y=\frac{\log(x+6)}{\log(2)}-5 \end{gathered}$

step 3

Let

$f^{(-1)}(x)=y$

therefore

the inverse function is

$f^{(-1)}(x)=\frac{\log(x+6)}{\log(2)}-5$

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1 year ago