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3 years ago

Reflections. Please help me

Mathematics

1 answer:

Marat540 [252]3 years ago

8 0

Answer:

see the attachments

Step-by-step explanation:

I suggested to you on a different occasion that by using tracing paper or tissue paper, you could make a copy of the image that you could move in the desired way to find its new location.

Here, the first attachment shows the figure being drawn on a piece of tissue with the line of reflection and the axes origin also shown.

The second attachment shows the tissue flopped over and the origin and line of reflection aligned with their previous locations. The new location of the figure is fairly obvious.

For reflection, any point that was some distance from the line on one side will be reflected to the same distance from the line on the other side. The distance is measured perpendicular to the line.

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Comment on "the work"

I only have a badly focused image of your original worksheet to work from, but even that is sufficient to illustrate the process and the result. It took longer to make and edit the photos than to do the drawing necessary to find the answer.

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4 years ago

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Multiple Roots
Standard Form: ax² + bx + c = 0
Quadratic Formula: $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Algebra II

Imaginary Root i = √-1

Step-by-step explanation:

Step 1: Define

Identify

2x² + x + 67 = 0

a = 2

b = 1

c = 67

Step 2: Solve for x

Substitute in variables [Quadratic Formula]: $\displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2(1)}$
Multiply: $\displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2}$
[√Radical] Evaluate exponents: $\displaystyle x=\frac{-1 \pm \sqrt{1-4(2)(67)}}{2}$
[√Radical] Multiply: $\displaystyle x=\frac{-1 \pm \sqrt{1-536}}{2}$
[√Radical] Subtract: $\displaystyle x=\frac{-1 \pm \sqrt{-535}}{2}$
[√Radical] Simplify: $\displaystyle x=\frac{-1 \pm i\sqrt{535}}{2}$

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