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kodGreya [7K]
2 years ago
12

The watermelon bought by Peter is 3 times as heavy as the papaya bought by Paul. If the watermelon bought by Peter has a mass of

4.2 kg, what is the mass of the papaya?​
Mathematics
2 answers:
Juli2301 [7.4K]2 years ago
5 0

Answer:

12.6

Step-by-step explanation:

4.2×3=12.6

I hope this answer can help you

Brrunno [24]2 years ago
4 0

Answer:

<u>1.4 kg</u>

Step-by-step explanation:

Given :

⇒ Watermelon (Peter) = 3 × Papaya (Paul)

⇒ Watermelon (Peter) = 4.2 kg

=============================================================

Solving :

⇒ 4.2 = 3 × Papaya (Paul)

⇒ Papaya (Paul) = <u>1.4 kg</u>

<u></u>

Therefore the mass of the papaya is 1.4 kg.

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Answer:

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Step-by-step explanation:

The shaded region is the difference of areas of semicircle AB and triangle ABC.

ABC is right triangle as AB is diameter, so C is right angle.

<u>Area of semicircle:</u>

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<u>Area of triangle:</u>

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<u>Shaded are is:</u>

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Step-by-step explanation:

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3 years ago
Factorize x3 -2x2 -5x+6
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2 years ago
A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.
Verdich [7]

Answer:

2a - 3b + 4c = 1

Step-by-step explanation:

Given

a^2 + b^2 + c^2 = 2(a - b - c) - 3

Required

Determine 2a - 3b + 4c

a^2 + b^2 + c^2 = 2(a - b - c) - 3

Open bracket

a^2 + b^2 + c^2 = 2a - 2b - 2c - 3

Equate the equation to 0

a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0

Express 3 as 1 + 1 + 1

a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0

Collect like terms

a^2 - 2a + 1 + b^2 + 2b + 1 + c^2  + 2c + 1 = 0

Group each terms

(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

Factorize (starting with the first bracket)

(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1)^2) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1)^2) + (b^2 + b+b + 1) + (c^2  + 2c + 1) = 0

((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)(b + 1)) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)^2) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)^2) + (c^2  + c+c + 1) = 0

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By comparison

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Take square root of both sides

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Add 1 to both sides

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Subtract 1 from both sides

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2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)

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