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3 years ago

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On a coordinate plane, triangle A has points (negative 2, 0), (negative 6, 0), (negative 4, 4). Triangle A prime has points (1,

0), (3, 0), (2, negative 2).
Which describes the sequence of transformations used to obtain the image (triangle A’) from the pre-image (triangle A)?
dilation by a scale factor of 2, then rotation of 180° about the origin
dilation by a scale factor of 2, then a reflection over the x-axis
dilation by a scale factor of 1/2, then a rotation of 180° about the origin
dilation by a scale factor of 1/2. then a reflection over the x-axis

2 answers:

Phantasy [73]3 years ago

6 0

Answer:

Step-by-step explanation:

B,D,E

Daniel [21]3 years ago

4 0

Answer:

c. dilation by a scale factor of 1/2, then a rotation of 180° about the origin

Step-by-step explanation:

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A small radio transmitter broadcasts in a 30 mile radius. If you drive along a straight line from a city 33 miles north of the t

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34.26miles from 49.58miles of the drive (69.11% of the drive).

Step-by-step explanation:

A problem of Analytic Geometry. This question can be solved using the resulting values from equaling the equations of the line (for the drive) and circle (with radius equals 30miles for radio transmitter broadcasting), and calculating the total distances from coincident points between circle and line, and total drive.

A graph is attached showing part of the circle and line with coincident points.

<h3>Line's Equation</h3>

Assuming transmitter is located (0, 0). From a city 33 miles north of the transmitter (0, 33) to a second city 37 miles east of the transmitter (37, 0).

Then, the equation for the line is (taking the two points from the start to the end):

$\\ y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}*(x-x_{1})$

$\\ y-33=\frac{0-33}{37-0}*(x-0)}$

$\\ y-33=\frac{-33}{37}*x}$

$\\ y=\frac{-33}{37}*x} + 33$ [1]

<h3>Circle's Equation</h3>

The equation for the circle whose center (0, 0) is:

$\\ (x-0)^2 + (y-0)^2 = 30^2$

$\\ x^2 + y^2 = 30^2$

$\\ y^2 = 30^2 - x^2$

$\\ y = \frac{+}{-}\sqrt{30^2- x^2}$ [2]

Equaling [1] and [2], to determine the points where starting to receive the radio transmitter signals to the point where finishing those signals:

$\\ \frac{-33}{37}*x + 33 = \sqrt{30^2- x^2}$

Solving this equation for <em>x</em>, we have two solutions for it (from <em>WolframAlpha</em>):

$\\ x_{1} = 40293/2458 - (111 \sqrt(80151))/2458 \approx 3.60775$

$\\ x_{2} = 40293/2458 + (111 \sqrt(80151))/2458 \approx 29.1774$

Then, using [1], the corresponding values for <em>y</em> are:

$\\ y_{1}=\frac{-33}{37}*(3.60775) \approx 29.78$

$\\ y_{2}=\frac{-33}{37}*(29.1774) \approx 6.97$

So,

$\\ (x_{1}=3.61, y_{1}=29.78)$

$\\ (x_{2}=29.18, y_{2}=6.97)$

Well, the distance of the drive is:

$\\ d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$

$\\ d = \sqrt{(29.18-3.61)^2 + (6.97-29.78)^2}$

$\\ d \approx 34.2654$ miles.

The total distance traveled is:

$\\ d = \sqrt{(37-0)^2 + (0-33)^2}$

$\\ d \approx 49.5782$ miles

Thus, during the drive, the signal of the radio transmitter was picked up for 34.26miles from the total of the 49.58 traveled, that is, a fraction equivalent to $\\ \frac{34.2654}{49.5782} \approx 0.6911$ or 69.11% of the drive.

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