To find the two other decimals that Holland could have used, we start by using the information provided:
- One of the decimals is 0.34.
- The sum of all three decimals is 2.
We can set up the equation to find the sum of the two other decimals. Let \( x \) and \( y \) be the other two decimals. We can express this relationship mathematically as:
\[ 0.34 + x + y = 2 \]
To isolate \( x + y \), we subtract 0.34 from both sides:
\[ x + y = 2 - 0.34 \]
Calculating the right side:
\[ 2 - 0.34 = 1.66 \]
Now we know that the sum of the two other decimals \( x \) and \( y \) must be:
\[ x + y = 1.66 \]
There are many pairs of decimals that can add up to 1.66. Some possible pairs include:
- \( x = 0.80 \) and \( y = 0.86 \) (since \( 0.80 + 0.86 = 1.66 \))
- \( x = 1.00 \) and \( y = 0.66 \) (since \( 1.00 + 0.66 = 1.66 \))
- \( x = 0.50 \) and \( y = 1.16 \) (since \( 0.50 + 1.16 = 1.66 \))
Any combination of decimals that satisfies \( x + y = 1.66 \) is valid.
Thus, Holland could have used, for example, the decimals \( 1.00 \) and \( 0.66 \) or any of the other pairs listed above to achieve a sum of 2 when added to 0.34.
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