Which of the following is not a rigid transformation?

2 answers:

8 0

Dilation, because the other transformations don’t manipulate size, but rather change the placement or making it mirrored

steposvetlana [31]3 years ago

5 0

The answer is A. Dilation

A dilation changes the actual size of the shape. All other transformations only move the shape or flip it.

hope this helps !

You might be interested in

2/4 + 3/9 in Modeling Fractions?

horrorfan [7]

Answer: 13/12

Step-by-step explanation:

Reduce the fraction 3/9 to the lowest terms by extracting and canceling out 3.

The least common multiple of 4 and 3 is 12. Convert 3/4 and 1/3 to fractions with denominator 12.

Since 9/12 and 4/12 have the same denominator, add them by adding their numerators.

Add 9 and 4 to get 13.

5 0

2 years ago

Help me please and I'm going to change my avatar in brainly to ice kachang babe

Finger [1]

Answer:

Q1: $1.2 Q2: 5 pens

Step-by-step explanation:

4 0

2 years ago

Find x+y. please help !!

Talja [164]

Answer:

use a calculator du bro 56

Step-by-step explanation:

5 0

3 years ago

Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4

steposvetlana [31]

$f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}$

$\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}$

6 0

3 years ago

Read 2 more answers

Marla received a $15 tip on a meal that cost $75.

stealth61 [152]

75x0.20=15 answer is 20%

3 0

2 years ago