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lara31 [8.8K]
4 years ago
15

Triangle ABC was transformed using the rule (x, y) → (–y, x). The vertices of the triangles are shown. A (–1, 1) B (1, 1) C (1,

4) A' (–1, –1) B' (–1, 1) C' (–4, 1) Which best describes the transformation? The transformation was a 90° rotation about the origin. The transformation was a 180° rotation about the origin. The transformation was a 270° rotation about the origin. The transformation was a 360° rotation about the origin.

Mathematics
2 answers:
yaroslaw [1]4 years ago
5 0

Answer:

the transformation was 90° at the origin

Step-by-step explanation:

the transformation was 90° at the origin

Vladimir [108]4 years ago
3 0

Answer:

A

Step-by-step explanation:

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Chang is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices.
Elena L [17]

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35

Step-by-step explanation:

86-65=21$ (initial fee diffrence)

21/0.6= 35

for 35 m company B charge less than A unless if renter want to drive further 35m so company A present a better offer.

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For the picture of the kite below, how would you prove that triangle BCE is congruent to triangle DCE if ∠1 is congruent to ∠2 a
aleksandrvk [35]

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6 0
3 years ago
What is the difference between quanitative and qualititave?
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'Quality' and 'quantity' are easy ways to remember  it :)
3 0
3 years ago
A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is
Brums [2.3K]

Answer:

\frac{259}{54}\text{ or }4.8\text{feet}

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one \frac{1}{6} as high as the preceding one.

TO FIND: If this ball is dropped from a height of 12 feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces \frac{1}{6} of preceding one and comes down the same distance.

When the ball is dropped from 12 feet height

after first hit =12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4

new height =12\times\frac{1}{6}=2\text{feet}

after second hit =2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}

new height =2\times\frac{1}{6}=\frac{1}{3}\text{ feet}

after third hit  =\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}

new height =\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}

after fourth hit =\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}

adding all distance =4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}

                                 =\frac{259}{54} feet

Hence the ball will travel   \frac{259}{54} feet before it hits the surface fifth time.

                                 

4 0
4 years ago
Is my answer correct?
BARSIC [14]

Answer:

It would be the first.

Step-by-step explanation:

a1 = -1

a2 = -1 + 0.5 = -0.5

a3 = -0.5 + 0.5 = 0

etc.

3 0
4 years ago
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