Question

The Statue of Liberty weighs about 2.04 x 105 kg. The Washington Monument weighs about 9.07 x 107 kg. About how many more kilograms does the Washington Monument weigh than the Statue of Liberty?

Answer 1

Answer:

$9.05 * 10^{7} kg$

Step-by-step explanation:

Given

$Statue\ of\ Liberty\ =\ 2.04 * 10^5 kg$

$Washington\ Monument\ =\ 9.07 * 10^7 kg$

To get the number of kilograms the Washington Monument weighs more than the Statue of Liberty, we simply calculate the difference;

Difference = Washington Monument - Statue of Liberty

$Difference = \ 9.07 * 10^7 kg - \ 2.04 * 10^5 kg$

Expand $10^7$

$Difference = \ 9.07 * 10^{5+2} kg - \ 2.04 * 10^5 kg$

$Difference = \ 9.07 * 10^{5} * 10^2 kg - \ 2.04 * 10^5 kg$

Take out the common factor in the above expression

$Difference = 10^{5}\ (\ 9.07 * 10^2 kg - \ 2.04\ kg)$

$10^2 = 100;$ so we have

$Difference = 10^{5}\ (\ 9.07 * 100 kg - \ 2.04\ kg)$

$Difference = 10^{5}\ (\ 907 kg - \ 2.04\ kg)$

$Difference = 10^{5}\ (904.96\ kg)$

Open bracket

$Difference =904.96 * 10^{5} kg$

Write in standard form

$Difference =9.0496 * 10^2 * 10^{5} kg$

$Difference =9.0496 * 10^7 kg$

$Difference =9.05 * 10^7 kg$ (Approximated)

Hence, he Washington Monument weigh more than the Statue of Liberty by about $9.05 * 10^{7} kg$