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SOVA2 [1]
3 years ago
10

Which transformation can NOT be used to prove that ABC is congruent to DEF?

Mathematics
2 answers:
Murrr4er [49]3 years ago
8 0

Answer:dialation

Step-by-step explanation:

Alexxx [7]3 years ago
7 0

Answer: it is dilation

Step-by-step explanation: i took the exam

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Nancy wrote the greatest number that can be made using each of these digits exactly once 5 3 4 9 8 1
lora16 [44]
985431 is your answer may man/Girl no prob attending ya
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3 years ago
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There are 32 students in a class.Nine of those students are women.What percent are men?(round nearest tenth)
Stels [109]
Hey there!
Find out how many men there are by subtracting the number of woman by the total number of students
32-9= 23
23 students are men
To find the percent just simply multiply. *cross multiply
23/32 *100= 71.87
100*23= 2300
2300/32= 17.875

17.875 rounds to 17.9

71.9% are men
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What is the area of the composite figure?
sasho [114]

Answer:

59.5

Step-by-step explanation:

(7x7) + (7x3)/2

8 0
3 years ago
Select all the ordered pairs that satisfy the function y = 3x -4.
aksik [14]

Answer:

B and D

Step-by-step explanation:

plug in the x and y for the x and y in the equation

if the two sides come out equal to eachother they satify the function

if they dont well they dont satisfy the function

hope this helps <3

4 0
3 years ago
in a five character password the first two characters must be digits and the last three characters must be letters if no charact
Vilka [71]

Answer:

1,404,000 unique passwords are possible.

Step-by-step explanation:

The order in which the letters and the digits are is important(AB is a different password than BA), which means that the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

P_{(n,x)} = \frac{n!}{(n-x)!}

In this question:

2 digits from a set of 10(there are 10 possible digits, 0-9).

3 characters from a set of 26. So

P_{10,2}P_{26,3} = \frac{10!}{8!} \times \frac{26!}{23!} = 10*9*26*25*24 = 1404000

1,404,000 unique passwords are possible.

5 0
3 years ago
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