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3 years ago

Reflection of a line

Mathematics

1 answer:

MakcuM [25]3 years ago

3 0

Hello!

Since this is a reflection, then the new figure has to be congruent since we're not changing the size or the shape.

Now, (x, y) is located in the First Quadrant, whereas (x, -y) is located in the Fourth Quadrant. If we're going from the First Quadrant to the Fourth Quadrant, then this has to be a reflection across the x-axis.

The correct answer is "Yes, ΔP'Q'R is a reflection of ΔPQR over the x-axis." :)

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4 years ago

Solve for g. Show your work! g+7 4=15

Varvara68 [4.7K]

Answer:

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Step-by-step explanation:

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3 years ago

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4 years ago

P= 2L +2W, solve for W

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6 0

3 years ago

No one has answered this one

krek1111 [17]

Answer:

$\displaystyle x = \frac{25\sqrt{2}}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

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

Trigonometry

[Right Triangles Only] SOHCAHTOA
[Right Triangles Only] cosθ = adjacent over hypotenuse

Step-by-step explanation:

Step 1: Define

Identify variables

Angle θ = 45°

Adjacent Leg = 25

Hypotenuse = x

Step 2: Solve for x

Substitute in variables [cosine]: $\displaystyle cos(45^\circ) = \frac{x}{25}$
[Multiplication Property of Equality] Multiply 25 on both sides: $\displaystyle 25cos(45^\circ) = x$
Rewrite: $\displaystyle x = 25cos(45^\circ)$
Evaluate: $\displaystyle x = \frac{25\sqrt{2}}{2}$

7 0

3 years ago