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ivann1987 [24]
3 years ago
13

The sum of three and a number is 24

Mathematics
2 answers:
Olin [163]3 years ago
7 0

Answer:

21

Step-by-step explanation:

let the number be 'x' then,

according to the question,

or, 3 + x = 24

or, x = 24 - 3

therefore , x = 21 ans

Whitepunk [10]3 years ago
4 0

Answer:

Step-by-step explanation:

whenever you see these questions, just subtract the first number, 3, from the total number. Doing this, 24 - 3 = 21

The number is 21

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On a blueprint, the scale indicates that 7 cm represent 16 ft. What is the length of a room that is 9.8 cm long and 5 cm wide on
Ira Lisetskai [31]
The room is 22.4 feet long

7cm=16ft
9.8cm=?ft

  7cm       9.8cm
______=______
  16ft          x ft

Cross multiply
7x=156.8
divide both sides by 7
x=22.4 ft


3 0
3 years ago
Read 2 more answers
Help me with this thanks for anyone who helps! :D
Kobotan [32]

Answer:

x= 90-60

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Step-by-step explanation:

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2 years ago
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If a = 12, b = 30, and c = 22, find the area of triangle ABC to the nearest tenth.
UNO [17]
7900 should be the answer, is there any options?

4 0
3 years ago
What is the area of the shaded region?
Mekhanik [1.2K]

9514 1404 393

Answer:

  434 -49π ≈ 280.1 cm²

Step-by-step explanation:

The shaded area is the difference between the enclosing rectangle area and the circle area.

The rectangle is 14 cm high and 31 cm wide, so has an area of ...

  A = WH

  A = (31 cm)(14 cm) = 434 cm²

The circle area is given by ...

  A = πr²

  A = π(7 cm)² = 49π cm²

__

The shaded area is the difference of these, so is ...

  shaded area = rectangle - circle

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4 0
2 years ago
Determine the zeroes of the polynomial
Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)}  \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)}  -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\

\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

8 0
3 years ago
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