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Drupady [299]
3 years ago
8

2.7x -9 = x +7.5 need the answer can't find solution.

Mathematics
2 answers:
Montano1993 [528]3 years ago
6 0

Answer:

x = 9.7

Step-by-step explanation:

2.7x -9 = x +7.5

2.7x - x = 7.5 + 9

1.7x = 16.5

x = 16.5/1.7

x = 9.7

andrew11 [14]3 years ago
6 0

Answer:

x = 9.70588

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

2.7x - 9 = x + 7.5

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Subtract <em>x</em> on both sides:                    1.7x - 9 = 7.5
  2. Add 9 on both sides:                           1.7x = 16.5
  3. Divide 1.7 on both sides:                     x = 9.70588

<u>Step 3: Check</u>

<em>Substitute in x into the original equation to verify it's a solution.</em>

  1. Substitute in <em>x</em>:                    2.7(9.70588) - 9 = 9.70588 + 7.5
  2. Multiply:                               26.2059 - 9 = 9.70588 + 7.5
  3. Subtract/Add:                      17.2059 = 17.2059

Here we see that 17.2059 does indeed equal 17.2059.

∴ x = 9.70588 is a solution of the equation.

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Answer:

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30 / (1/10) = 300

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Step-by-step explanation:

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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.

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