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2 years ago

Which best describes a transformation that preserves the size, shape, and angles of an object?

Mathematics

2 answers:

Alex787 [66]2 years ago

5 0

Isometry is what you just described.

kari74 [83]2 years ago

4 0

Answer:

The answer is isometry.

Step-by-step explanation:

Isometry best describes a transformation that preserves the size, shape, and angles of an object. An isometric transformation is a shape preserving transformation in a plane or in space.

Isometries include rotation, reflection, translation, glides etc.

Though options are not given, but isometry is the correct answer.

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What is the solution? y = 3x - 5 y = 1\3*+3

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

y = 3x - 5

y = 1/3x + 3

3x - 5 = 1/3x + 3....multiply everything by 3 to get rid of the fraction

9x - 15 = x + 9

9x - x = 9 + 15

8x = 24

x = 24/8

x = 3

y = 3x - 5

y = 3(3) - 5

y = 9 - 5

y = 4

solution is : x = 3 and y = 4...or (3,4) <==

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2 years ago

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2 years ago

Which of the following is a solution of x2 + 6x = −18?

Rus_ich [418]

X= 9/4 i think i dont know

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2 years ago

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Neporo4naja [7]

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1 year ago

Change the subject of the formula L = v 4kt - p to k.

Romashka [77]

Answer:

$\boxed{k = \frac{L^2 + p}{4t}}$

General Formulas and Concepts:

Algebra I

Basic Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle L = \sqrt{4kt - p}$

Step 2: Solve for k
We can use equality properties to help us rewrite the equation to get k as our subject:

Let's first square both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\\end{aligned}$

Next, add p to both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\\end{aligned}$

Next, divide 4t by both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\& \rightarrow \frac{L^2 + p}{4t} = k \\\end{aligned}$

We can rewrite the new equation by swapping sides to obtain our final expression:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow \boxed{k = \frac{L^2 + p}{4t}}\end{aligned}$

∴ we have changed the subject of the formula.

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Learn more about Algebra I: brainly.com/question/27698547

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Topic: Algebra I

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1 year ago