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Alexus [3.1K]
3 years ago
11

Square ABCD has coordinates A(−10, 5) , B(10, 5), C(10, 0) , and D(−10, 0) . Square A'B'C'D' has coordinates A′(−10,−5),B′(10,−5

) , C'(10, 0) , and D′(−10, 0) . Square A"B"C"D" has coordinates A′′(−2, −1), B′′(2, −1) , C"(2, 0) , and D′′(−2, 0) .
Which transformations describe why squares ABCD and A"B"C"D" are similar?

​ A).Square ABCD ​ was reflected across the y-axis and then dilated by a scale factor of 5 to form ​ square A"B"C"D" ​

​ B).Square ABCD ​ was dilated by a scale factor of 5 and then rotated 90° counterclockwise to form ​ square A"B"C"D" ​ .

C).​ Square ABCD ​ was reflected across the x-axis and then dilated by a scale factor of 1/5 to form square A"B"C"D" .

D).​ Square ABCD ​ was dilated by a scale factor of 1/5 and then rotated 90° counterclockwise to form ​ square A"B"C"D" ​
Mathematics
1 answer:
neonofarm [45]3 years ago
8 0
The answer would either be A or D
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8-10 deal with the ambiguous SSA case. for each find all possible solutions and sketch the triangle in each case.
Artist 52 [7]
<span>For problem 8:

sin A/a sinB/b = sinC/c 

so sin50/15 = sinB/12 

0.766/15 =sinB/12 

0.051066 = sin B/12 

sin B = 0.6128 
B = 37.79 degrees 

sum of angles must equal 180 deg therefore C = 180-50-37.79 = 92.21 degrees and .766/15 = sin92.21/c 

.051066= 0.99925/c 
c = .99925/0.051066 = 19.567 

Problems 9 and 10 can also be done with the same method as problem 8.

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Need help with stats!
Brut [27]

Answer:

a) 1,440 ways

b) 59,280 or 64,000

Step-by-step explanation:

a) Aircraft boarding.

8 people, 2 in first class, boarding first, then 8 economy class.

The 2 people in first class board first, but they can board as AB or BA... so 2 ways here.

For the 6 economy class passengers, we have a permutation of 6 out of 6, so 720, as follows:

P(6,6) = \frac{6!}{(6 - 6)!} = 6! = 720

Since the two are independent, we multiply them to have a global number of ways: 2 * 720 = 1,440 different ways for the 8 passengers to board that plane.

b) combination lock.

Here we do have a little problem... the question doesn't specify if the 3 numbers are different numbers of not.  So, we'll calculate both:

Numbers go from 1 to 40 inclusively... so 40 possibilities.

Normally, in a combination lock, the numbers are different, so let's start with that one:

First number: 40 options available

Second number: 39 options available (cannot take the first one again)

Third number: 38 different options (can't take First or Second number again)

Overall, we then have 40 * 39 * 38 = 59,280 different lock combinations.

If we can pick pick the same number twice:

First number: 40 options available

Second number: 40 options available

Third number: 40 options available

Overall 40 * 40 * 40 = 64,000 different lock combinations

8 0
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