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zhannawk [14.2K]
3 years ago
15

PLZ HELP

Mathematics
1 answer:
maria [59]3 years ago
5 0

Answer:

Step-by-step explanation:

Following changes will be there when the figure is transformed by the given rules.

1). Rule for transformation has been given as,

   (x, y) → (x, -y)

   Reflection across x axis.

2). (x, y) → (-x, -y)

   Rotation of 180° about the origin.

3). (x, y) → (x - 4, y)

   Shifted 4 units left horizontally.

4). (x, y) → (x, y + 3)

   Shifted vertically up by 3 units

5). (x, y) → (x - 1, y + 4)

   Shifted 1 units left horizontally and 4 units up vertically.

6). (x, y) → (4x, 4y)

   Dilated by 4 units.

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The answer is A, 13, -13.

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Which functions are symmetric with respect to the origin?
Damm [24]

Answer:

The correct option is;

y = arcsinx and y = arctanx

Step-by-step explanation:

The given options are;

1) y = arcsinx and y = arccosx

Here, we have at the origin, where x = 0,  arccosx ≈ 1.57 while arcsinx = 0

Therefore arccosx does not intersect arcsinx at the origin for it to be symmetrical to arcsinx or the origin

2)  y = arccosxy and y = arctanx

Here arctanx = 0 when x = 0 and arcos x = 1.57 when x = 0 therefore, they are not symmetrical

3) y = arctanx and y = arccotx

Similarly, At x = 0, arccotx = 1.57 therefore, they are not symmetrical

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Help me with these math questions....
Nady [450]

Answer: cotθ

<u>Step-by-step explanation:</u>

 tanθ * cos²θ * csc²θ

=  \dfrac{sin\theta}{cos\theta} * \dfrac{cos\theta*cos\theta}{} *\dfrac{1}{sin\theta*sin\theta}

= \dfrac{cos\theta}{sin\theta}

= cotθ

Answer: B

<u>Step-by-step explanation:</u>

The parent graph is y = x²

The new graph y = -x² + 3 should have the following:

  • reflection over the x-axis
  • vertical shift up 3 units

Answers:

  • a. Quadrant II
  • b. negative
  • c. \dfrac{\pi}{6}
  • d. C
  • e.-\dfrac{\sqrt{3}}{3}

<u>Explanation:</u>

\dfrac{17\pi}{6} - \dfrac{12\pi}{6} = \dfrac{5\pi}{6}

a) Quadrant 2 is: \dfrac{\pi}{2} < \theta < \pi

b) In Quadrant 2, cos is negative and sin is positive, so tan is negative

c) \pi-\dfrac{5\pi}{6} = \dfrac{\pi}{6}

d) the reference line is above the x-axis so it is negative -->  -tan\dfrac{\pi}{6}

e) tan(\dfrac{5\pi}{6})=\dfrac{1}{-\sqrt{3}}=-\dfrac{\sqrt{3}}{3}


4 0
3 years ago
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