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Feliz [49]
2 years ago
14

30 6aın yüzde kaçıdır​

Mathematics
1 answer:
Julli [10]2 years ago
6 0

Answer:

% 20

Step-by-step explanation:

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Answer:

What do u need help with I don't get what u need help with

Step-by-step explanation:

4 0
2 years ago
Convert the following into millilitres a)13 litres​
skelet666 [1.2K]

Answer:

13,000ml

Step-by-step explanation:

1000ml = 1L.

We can rewrite this forumla as:

yL x 1000=ml. y=the number of liters.

In this problem, y=13. Thus:

13Lx1000= 13,000ml

I hope this helps!

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3 years ago
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5 0
3 years ago
PLEASE HELP ME!!
Grace [21]
Take the derivative:
g’(x) = 12x^3 - 24x^2

Set equal to zero and solve:
0 = 12x^3 - 24x^2
0 = 12x^2 (x - 2)

x = 0 or x = 2

Plug back into original
g(0) = 3(0^4) - 8(0^3)
g(0) = 0 - 0
g(0) = 0

g(2) = 3(2^2) - 8(2^3)
g(2) = 3(4) - 8(8)
g(2) = 12 - 64
g(2) = -52

There is an absolute max at (0,0) or when x = 0
5 0
2 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
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