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Norma-Jean [14]
2 years ago
13

Which pair of angles is congruent by ASA?

Mathematics
2 answers:
pogonyaev2 years ago
8 0

Answer:

C

Step-by-step explanation:

Leya [2.2K]2 years ago
5 0

Answer:

its c

Step-by-step explanation:

because yeah its c

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Answer the picture below.
inysia [295]

Answer:

Step-by-step explanation:

function: all the points

note: x intercept is the domain, y intercept is the range

domain must paired with exact range

the domain can be more than 1, but range must be 1.

domain : 3, 5, 0 , -1, -3, -5

range: 1, 3, 4 , -3(there is 2 but we write it one) , -4

and done!

thank you!

4 0
2 years ago
A manufacturing plant earned \$80$80dollar sign, 80 per man-hour of labor when it opened. Each year, the plant earns an addition
vagabundo [1.1K]

Answer:

A(t)=80(1.05)^t

Step-by-step explanation:

The manufacturing plant earned $80 per man-hour of labor when it opened

The plant earns an additional 5% for every additional man-hour t.

This can be modeled using the function:

A(t)=P(1+r)^t

where Initial Amount Earned, P=80

Rate of Increase, r=5%=0.05

Therefore, the function that models the amount earned at any time t is:

A(t)=80(1+0.05)^t\\A(t)=80(1.05)^t

3 0
3 years ago
There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl
Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

6 0
3 years ago
A standard number cube is tossed. Find the following probability. ​P(greater than 4 or odd​)
lorasvet [3.4K]
A standard cube has 6 possible outcomes. But since we want to find P(4+ or odd), the outcomes become more narrow.
So:
6 outcomes -> 1, 2, 3, 4, 5, 6

But we want to find 4+
So:
4+ = 4, 5, 6

We also want to find all odd possibilities.
So:
odd = 1, 3, 5

As you can see, there is a 5 in both the 4+ and the odd numbers so 5 will only be counted once.
P(4+ or odd) = 1, 3, 4, 5, 6

So the probability of having an outcome that is P(4+ or odd) is:
5/6 or 0.8333 or 83.33%

Hope I helped!
8 0
2 years ago
Read 2 more answers
Which statements about a square are always true?
blondinia [14]
All of the answers are correct except the last one
6 0
2 years ago
Read 2 more answers
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