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2 years ago

Using The diagram, name two pairs of adjacent angles and two pairs of vertical angles.

Mathematics

1 answer:

Crank2 years ago

3 0

Answer:

1. adjacent: ∠ACB and ∠BCD.

vertical: ∠ACB and ∠ECD

Step-by-step explanation:

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Pre image ABCD was dilated to produce image A’B’C’D why is the scale factor from the pre image? Enter your answer in the box as

shepuryov [24]

The scale factor of the dilation from ABCD to A′B′C′D′ is 3.

Step-by-step explanation:

Step 1:

In the pre-image ABCD, the length of one of the sides is given as 14 units.

For the other shape A′B′C′D′, the same side as the previous shape is given as 8 units.

Step 2:

To determine the scale factor, we divide the measurement after scaling by the same measurement before scaling.

In this case, it is the given length of the sides CD and C′D′.

So the scale factor = $\frac{C^{1} D^{1} }{CD} = \frac{8}{14} = \frac{4}{7} .$

So the shape ABCD is dilated by a scale factor of $\frac{4}{7}$ to produce the shape A′B′C′D′.

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3 years ago

What is the solution to the given inequality?

xenn [34]

Answer: x ≤3

Step-by-step explanation:

Your answer is correct

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2 years ago

How do I use substitution to solve

nekit [7.7K]

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3 years ago

Read 2 more answers

There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit

Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a $\frac{3}{11}$ chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is $\frac{8}{11}$ , and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

$\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}$

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is $\frac{2}{11}$ , and the probability of picking the second red ball is $\frac{1}{10}$ and the probability of picking the green ball is $\frac{1}{9}$ . We want to multiply thm again, as per the multiplication rule like the last problem.

$\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}$

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3 years ago

What is the answer to this question

il63 [147K]

The ratio of the shortest sides is 12/5. Multiplying the other two sides of the smaller triangle by that ratio gives 15 and 18, so the ratios of all sides are the same. The triangles are similar.

ABC ~ ZYX

The largest/smallest scale factor is 12/5 = 2.4.

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3 years ago

Using The diagram, name two pairs of adjacent angles and two pairs of vertical angles.​

Using The diagram, name two pairs of adjacent angles and two pairs of vertical angles.