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mafiozo [28]
3 years ago
8

Which sequence of transformations creates a similar, but not congruent, triangle? A. Rotation and translation B. Reflection and

rotation C. Dilation and rotation D. Translation and reflection
Mathematics
2 answers:
Nat2105 [25]3 years ago
4 0
<span><u><em>The correct answer is: </em></u>
dilation and rotation.

<u><em>Explanation</em></u><span><u><em>: </em></u>
Rotations, reflections and translations are known as rigid transformations; this means they do not change the size or shape of a figure, they simply move it. These rigid transformations preserve congruence.

Dilation, however, are not rigid transformations, since they change the size of a shape. Dilation would not change the shape, just the size; the angle measures would be the same, and the ratio of corresponding sides would be equal to the scale factor used in the dilation. This would give us a similar, but not congruent, figure.</span></span>
Gala2k [10]3 years ago
3 0

Answer:the answer is dilation and reflection

Step-by-step explanation:

I did the question

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There are 92 students that equals 75 how many students are 100 percent
elena-s [515]

Answer:

122.67 or round it to 123

Step-by-step explanation:

i used the butterfly method.

92/75 = x/100.

i multiplied 92*100=9200

then i divided it with 75.

9200/75=x

so the answer is

x=122 or 123

4 0
3 years ago
Read 2 more answers
Is square root 1 minus sine squared theta = cos Θ true? If so, in which quadrants does angle Θ terminate?
Cloud [144]

Answer:

True; quadrants I & IV

Step-by-step explanation:

We know the relation between sine and cosine function which is given by

\sin^2 \theta +\cos^2 \theta = 1

Let us solve this equation for cosine function.

\cos^2 \theta = 1-\sin^2 \theta

Take square root both sides. When ever we take square root we need to write the solution in plus minus form

\sqrt{\cos^2 \theta}=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=-\sqrt{1-\sin^2 \theta}, \sqrt{1-\sin^2 \theta}

If Θ is in quadrants I and IV then the value will be positive and if Θ is in II and III quadrant then the value is negative.

Hence, if Θ is in quadrants I & IV, then we have

\cos \theta=\sqrt{1-\sin^2 \theta}

Thus, the correct option is: True; quadrants I & IV


6 0
4 years ago
Read 2 more answers
11 of 16
love history [14]

Answer:

Step-by-step explanation:

This is a geometric sequence; we know that because of the words "common ratio."  The general formula for a geometric sequence is

a(n) = a(1)*r^(n - 1), where r is the common ratio.

Here, a(1) = 1, r = 1/3, and so

a(1) = 1 (given)

a(2) = 1/3 (this is a(1) multiplied by 1/3)

a(3) = 1/9 (this is a(2) multiplied by 1/3)

a(4) = 1/27

a(5) = 1/81

8 0
3 years ago
Jack can paint 1 house in 3 days. Find Jack’s rate per day for painting houses.
Eduardwww [97]

Answer:

Jack can paint a certain room in 3 hours, lisa is 4 hours, and sue in 2 hours. ... Jack's rate is one-third of a job per hour, Lisa's is one-fourth of a job per hour and Sue's is one-half of a job per hour. If they all work together for t hours, one whole job is completed. t/3 + t/4 + t/2 = 1 .

Step-by-step explanation:

6 0
3 years ago
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. Waipio Rd in Honokaa, Hawaii, is one of the steepest roads in the world. It rises 450 ft over a horizontal distance of 1,000 f
Deffense [45]

Answer: 50%

Step-by-step explanation:

From the question, we're given the information that Waipio Rd in Honokaa, Hawaii, is one of the steepest roads in the world and that it rises 450 ft over a horizontal distance of 1,000 ft.

Therefore, the percentage grade of this hill will be calculated as the height over the length and this will be:

= 450/1000 × 100

= 45%

= 0.45

= 0.5 to nearest tenth

= 50% to nearest tenth.

8 0
3 years ago
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