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Georgia [21]
3 years ago
11

What is the image point of (-3,4)after the transformation D2∘R90∘ ​

Mathematics
1 answer:
zhenek [66]3 years ago
7 0

A composite transformation consists of two or more transformations with a procedure of starting from the rightmost transformation

The location of the image of the point (-3, 4), after the transformation D_2 \circ R90^{\circ} is <u>(-8, -6)</u>

Reason:

The given coordinate is (-3, 4)

The required transformation is D_2 \circ R90^{\circ}

Solution:

A rotation transformation of (x, y) 90° gives (-y, x)

Therefore;

The rotation of the point (-3, 4) 90° counterclockwise gives (-4, -3)

A transformation of D2 is a dilation by a scale factor of 2

Therefore; D2(-4, -3) gives 2×(-4, -3) = (-8, -6)

Which gives;

The image of (-3, 4), after the transformation  D_2 \circ R90^{\circ}  → (-8, -6)

Learn more about composite transformations here;

brainly.com/question/12366556

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A point H is 20m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the po
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Answer:

a. See Attachment 1

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c. HT = 31.1\ m

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Step-by-step explanation:

Calculating PT

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tan50 = \frac{OT}{20}

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Calculating OP;

We have to consider angle 30, distance OH and distance OP

The relationship between these parameters is;

tan30 = \frac{OP}{20}

Multiply both sides by 20

20 * tan30 = \frac{OP}{20} * 20

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11.548 = OP

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

Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

HT^2 = OT^2 + OH^2

Substitute 20 for OH and OT = 23.836

HT^2 = 20^2 + 23.836^2

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HT^2 = 968.154896

Take Square Root of both sides

HT = \sqrt{968.154896}

HT = 31.1\ m <em>(Approximated)</em>

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

Calculating the position of H

This is represented by OH

See Attachment 2

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

tan50 = \frac{OH}{OT}

Multiply both sides by OT

OT * tan50 = \frac{OH}{OT} * OT

OT * tan50 = {OH

OT * 1.1918 = OH

Substitute OT = 23.836

23.836 * 1.1918 = OH

28.4= OH

OH = 28.4\ m<em> (Approximated)</em>

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3 years ago
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