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2 years ago

Which two sets of angles are corresponding angles?

Mathematics

1 answer:

mojhsa [17]2 years ago

8 0

Answer:

∠a and ∠d; ∠b and ∠c

Step-by-step explanation:

The two triangles as similar triangles and the scale factor is 2 : 1

The sides measuring 6 and 3 are corresponding sides

The sides measeuring 8 and 4 are ocrreposnding sides

⇒ ∠a and ∠d are corresponding angles

The sides measuring 8 and 4 are corresponding sides

The sides measeuring 4 and 2 are ocrreposnding sides

⇒ ∠b and ∠c are corresponding angles

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Quadrilateral ABCD is a rectangle. If the coordinates of A

Neporo4naja [7]

<h3>Answer: A) -2</h3>

==========================================

Explanation:

Let's find the slope of line AB

m = (y2-y1)/(x2-x1)

m = (2-7)/(-6-4)

m = (-5)/(-10)

m = 1/2

The slope of line AB is 1/2

To get the slope of BC, we flip the fraction and the sign to get the perpendicular slope

1/2 becomes -2/1 or simply -2

----------

Slope of AB = 1/2

Slope of BC = -2

Notice how the two perpendicular slopes multiply to -1.

We know they are perpendicular because the four angles of any rectangle are always 90 degrees each.

5 0

2 years ago

Make n the subject of the formula t= square root of n+3/n

bazaltina [42]

Step-by-step explanation:

hi I have answered ur question

4 0

3 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

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Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$