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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?

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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Leno4ka [110]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

Factorize x3 -2x2 -5x+6

Advocard [28]

Answer: x³-2x²-5x-6=(x-1)*(x-3)*(x+2).

Step-by-step explanation:

$x^3-2x^2-5x+6=\\\\x^3-2x^2+x-x-5x+6=\\\\(x^3-2x^2+x)-6x+6=\\\\x*(x^2-2x+1)-6*(x-1)=\\\\x*(x-1)^2-6*(x-1)=\\\\(x-1)*(x*(x-1)-6)=\\\\(x-1)*(x^2-x-6)=\\\\(x-1)*(x^2-x-2x+2x-6)=\\\\(x-1)*(x^2-3x+2x-6)=\\\\(x-1)*(x*(x-3)+2*(x-3))=\\\\(x-1)*(x-3)*(x+2).$

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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)

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

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

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

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

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

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

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

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

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

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

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

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

Express 0 as 0 + 0 + 0

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

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

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

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

3 years ago

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Nastasia [14]

Hi, TRUEBOSS72!

The answer to your question I believe is $t(x) = 309.125^{x}$

-ASIAX ∵<span>Frequent Answerer</span>∴

P.S. If I am wrong about it please let me know.

3 0

3 years ago