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Which transformation of Figure A results in Figure A' ?

MariettaO [177]

Option B: a counterclockwise rotation of 90° about the origin

Explanation:

From the graph, we can see the coordinates of the figure A are (0,2), (-1,6) and (-4,4)

The coordinates of the figure A' are (-2,0), (-6,-1) and (-4,-4)

<u>Option B: a counterclockwise rotation of 90° about the origin </u>

The transformation rule for a coordinate to reflect a counterclockwise rotation of 90° about the origin is given by

$(x,y)\implies (-y,x)$

Let us substitute the coordinates of the figure A

Thus, we have,

$(0,2)\implies(-2,0)$

$(-1,6)\implies (-6,-1)$

$(-4,4)\implies(-4,-4)$

Thus, the resulting coordinates are equivalent to the coordinates of the figure A'.

Therefore, the figure is a counterclockwise rotation of 90° about the origin .

Hence, Option B is the correct answer.

3 0

3 years ago

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Multi-Step: <ABD and <BDE are supplementary. Find the measures of both angles.

Stella [2.4K]

Answer:

get it helps me with lI do not know the answer

Step-by-step explanation:

sorry

4 0

3 years ago

For the function g(x)=3-8(1/4)^2-x

Reika [66]

Using function concepts, it is found that:

a) The y-intercept is y = 2.5.

b) The horizontal asymptote is x = 3.

c) The function is decreasing.

d) The domain is $(-\infty,\infty)$ and the range is $(-\infty,3)$ .
e) The graph is given at the end of the answer.

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The given function is:

$g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}$

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Question a:

The y-intercept is g(0), thus:

$g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5$

The y-intercept is y = 2.5.

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

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3$

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$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty$

Thus, the horizontal asymptote is x = 3.

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Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

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Question d:

Exponential function has no restrictions in the domain, so it is all real values, that is $(-\infty,\infty)$ .
From the limits in item c, the range is: $(-\infty,3)$

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The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631

8 0

2 years ago

HELP!!! DUE IN AN HOUR!!

Ulleksa [173]

Answer:

measure 3 equals 121 degrees

Step-by-step explanation:

Measure 2 and 3 make a 180 angle and you subtract 59 from 180 and you get 121.

5 0

3 years ago

In a group discussion, which option is most clearly a clarifying question?

Mariulka [41]

Answer:

your answer should be d

Step-by-step explanation:

5 0

3 years ago

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