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weeeeeb [17]
3 years ago
6

The "HOPE PROBE " launched by Abu Dhabi weighed 1350 Kg along with the fuel it carried . If it uses 576 Kg of fuel during its jo

urney, can you find its present weight ?
Mathematics
1 answer:
dedylja [7]3 years ago
3 0

Answer: 774 kg

Step-by-step explanation:

Given

Dhabi weighted 1350 kg

It uses 576 kg of fuel during its journey

So, the remaining fuel is given by the difference of initial and used fuel present i.e.

\Rightarrow 1350-576=774\ kg

The present weight is 774 kg

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1.  The 5/65 can be simplified, the a³ moves to the denominator, and bc is canceled from the numerator and denominator.
\frac{5a^3 bc^2}{65 a^4 bc} = \frac{c}{13a}

2.  The 10/125 can be simplified, the a³ moves to the denominator, and bc is canceled from the numerator and denominator.
\frac{10a^3bc^2}{125a^4bc} = \frac{2c}{25a}

3.  The 65/5 simplifies, the a³ moves to the numerator, and bc is canceled from the numerator and denominator.
\frac{65a^4 bc^2}{5a^3 bc} = 13ac

4.  The 10/130 simplifies, the b² cancels out in the numerator and denominator, and the c^4 moves to the numerator.
\frac{10b^2 c^5}{130a^4b^2c^4} = \frac{c}{13a^4}

5.  The 10/130 simplifies, the b² cancels out in the numerator and denominator, the a^4 moves to the numerator, and the c^4 moves to the numerator.
\frac{10a^5b^2c^5}{130a^4b^2c^4} = \frac{ac}{13}


I believe the only expression that simplifies to \frac{c}{13a} is the first one.
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