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3 years ago

Which transformation gives the same result as a rotation of 180° around the origin

Mathematics

2 answers:

lana66690 [7]3 years ago

6 0

Answer:

Transformation gives the same result as a rotation of 180° around the origin, followed by a reflection over the y-axis is a reflection over the x-axis.

Hope this helps!

ValentinkaMS [17]3 years ago

5 0

Answer:

a reflection over the x-axis

Step-by-step explanation:

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<h2>Question :</h2>

$\tt \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

<h2>Answer :</h2>

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

<h2>Explanation :</h2>

$\tt : \implies \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)(x+2) + (x-2)(x-2)}{(x-2)(x+2)} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)^{2} + (x-2)^{2}}{(x-2)(x+2)} = \dfrac{5}{6}$

<u>Now, we know that</u> :

$\large \underline{\boxed{\bf{(a+b)^{2} = a^{2} + b^{2}+ 2ab}}}$
$\large \underline{\boxed{\bf{(a-b)^{2} = a^{2} + b^{2} - 2ab}}}$
$\large \underline{\boxed{\bf{(a+b)(a-b) = a^{2} - b^{2}}}}$

$\tt : \implies \dfrac{x^{2}+2^{2}+ 2 \times x \times 2 + x^{2}+2^{2} - 2 \times x \times 2 }{x^{2}-2^{2}} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2}+ 4 + \cancel{4x} + x^{2}+ 4 - \cancel{4x}}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2} + x^{2} + 4 + 4}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{2x^{2} + 8}{x^{2}-4} = \dfrac{5}{6}$

<u>By cross multiply</u> :

$\tt : \implies (2x^{2} + 8)6= 5(x^{2}-4)$

$\tt : \implies 12x^{2} + 48 = 5x^{2}-20$

$\tt : \implies 12x^{2} + 48 - 5x^{2} + 20 = 0$

$\tt : \implies 7x^{2} + 68 = 0$

$\tt : \implies 7x^{2} + 0x + 68 = 0$

<u>Now, by comparing with ax² + bx + c = 0, we have</u> :

a = 7
b = 0
c = 68

<u>By using quadratic formula</u> :

$\large \underline{\boxed{\bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}}$

$\tt : \implies x = \dfrac{-(0) \pm \sqrt{(0)^{2} - 4(7)(68)}}{2(7)}$

$\tt : \implies x = \dfrac{0 \pm \sqrt{0 - 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{- 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 17}}{14}$

$\tt : \implies x = \dfrac{\pm \cancel{2} \times 2\sqrt{7\times 17}}{\cancel{14}}$

$\tt : \implies x = \dfrac{\pm2\sqrt{119}}{7}$

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

Hence value of $\bf x =\dfrac{\pm 2\sqrt{119}}{7}$

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5. The distance between two given points (5,2) and (x,5) is 5 units. Find the possible values of x. *

kirza4 [7]

<h3>Answers: x = 1 and x = 9</h3>

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

Explanation:

We'll use the distance formula here. Rather than compute the distance d based on two points given, we'll go in reverse to use the given distance d to find what the coordinate must be to satisfy the conditions.

We're given that d = 5

The first point is $(x_1,y_1) = (5,2)$ and the second point has coordinates of $(x_2,y_2) = (x,5)$ where x is some real number.

We'll plug all this into the distance formula and solve for x.

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\5 = \sqrt{(5-x)^2+(2-5)^2}\\\\5 = \sqrt{(5-x)^2+(-3)^2}\\\\5 = \sqrt{(5-x)^2+9}\\\\\sqrt{(5-x)^2+9} = 5\\\\(5-x)^2+9 = 5^2\\\\(5-x)^2+9 = 25\\\\(5-x)^2 = 25-9\\\\(5-x)^2 = 16\\\\5-x = \pm\sqrt{16}\\\\5-x = 4 \ \text{ or } \ 5-x = -4\\\\-x = 4-5 \ \text{ or } \ -x = -4-5\\\\-x = -1 \ \text{ or } \ -x = -9\\\\x = 1 \ \text{ or } \ x = 9\\\\$

This means that if we had these three points

A = (5, 2)
B = (1, 5)
C = (9, 5)

Then segments AB and AC are each 5 units long.

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