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3 years ago

Which of the following steps were applied to abcd to obtain abcd

Mathematics

2 answers:

bulgar [2K]3 years ago

7 0

Answer:

Step-by-step explanation:

Move to the right 2 units,

Move down 2 units.

Damm [24]3 years ago

7 0

Answer:

Option (c) is correct.

Shift 2 units right and 3 units down.

Step-by-step explanation:

Given : A graph showing two quadrilaterals ABCD and A'B'C'D'

We have to choose the correct option from the given options that describes the steps applied to ABCD to obtain A'B'C'D'.

Translation is moving every coordinate of the image in same direction and in same distance.

Here, Coordinate of A is (3,5) and Coordinate of A' is (5,2)

Thus, To move A to A' we have shifted x coordinate 2 units to right and y coordinate 3 units down.

Similarly, we can check for each point

Consider Coordinate of B is (4,8) and Coordinate of B' is (6,5)

Thus, To move B to B' we have shifted x coordinate 2 units to right and y coordinate 3 units down.

Thus, The steps applied to ABCD to obtain A'B'C'D' is shifted x coordinate 2 units to right and y coordinate 3 units down.

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Assoli18 [71]

Answer: $6\sqrt{3}$

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

Explanation:

Method 1

We can use the pythagorean theorem to find x.

$a^2+b^2 = c^2\\\\6^2+x^2 = 12^2\\\\36+x^2 = 144\\\\x^2 = 144-36\\\\x^2 = 108\\\\x = \sqrt{108}\\\\x = \sqrt{36*3}\\\\x = \sqrt{36}*\sqrt{3}\\\\x = 6\sqrt{3}\\\\$

-----------------------------------

Method 2

Use the sine ratio to find x. You'll need a reference sheet or the unit circle, or simply memorize that sin(60) = sqrt(3)/2

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(60^{\circ}) = \frac{x}{12}\\\\\frac{\sqrt{3}}{2} = \frac{x}{12}\\\\x = 12*\frac{\sqrt{3}}{2}\\\\x = 6\sqrt{3}\\\\$

-----------------------------------

Method 3

Similar to the previous method, but we'll use tangent this time.

Use a reference sheet, unit circle, or memorize that tan(60) = sqrt(3)

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(60^{\circ}) = \frac{x}{6}\\\\\sqrt{3} = \frac{x}{6}\\\\x = 6\sqrt{3}\\\\$

-----------------------------------

Method 4

This is a 30-60-90 triangle. In other words, the angles are 30 degrees, 60 degrees, and 90 degrees.

Because of this special type of triangle, we know that the long leg is exactly sqrt(3) times that of the short leg.

$\text{long leg} = (\text{short leg})*\sqrt{3}\\\\x = 6\sqrt{3}\\\\$

The short leg is always opposite the smallest angle (30 degrees).

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