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BCAL-WORE CONNECTION The diagram shows the plans for the triangular park. How much longer is \( B C \) than \( A C \) ? \( \left(3 x^{2}-25 x-12\right) \) feet \( \left(3 x^{2}-5 x-12\right) \) feet \( \left(3 x^{2}+5 x-4\right) \) feet \( \left(3 x^{2}+25 x-4\right) \) feet DONE

Ask by Stephens Patel. in the United States
Feb 21,2025

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Answer

\( BC \) is shorter than \( AC \) by \( 20x \) feet.

Solution

To find out how much longer \( BC \) is than \( AC \), we need to subtract the length of \( AC \) from the length of \( BC \). Given: - Length of \( BC = 3x^2 - 25x - 12 \) feet - Length of \( AC = 3x^2 - 5x - 12 \) feet We can express the difference as follows: \[ BC - AC = (3x^2 - 25x - 12) - (3x^2 - 5x - 12) \] Now, let's simplify this expression step by step: 1. Distribute the negative sign across the second expression: \[ BC - AC = 3x^2 - 25x - 12 - 3x^2 + 5x + 12 \] 2. Combine like terms: - The \( 3x^2 \) terms cancel out: \[ BC - AC = -25x + 5x - 12 + 12 \] - This simplifies to: \[ BC - AC = -20x \] Thus, the difference in lengths is \( -20x \) feet. This means that \( AC \) is longer than \( BC \) by \( 20x \) feet. To summarize: - \( BC \) is shorter than \( AC \) by \( 20x \) feet.

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To find how much longer \( BC \) is than \( AC \), we need to subtract the expression for \( AC \) from the expression for \( BC \). This means we'll take: \[ BC - AC = (3x^2 - 25x - 12) - (3x^2 - 5x - 12) \] When you simplify this, you’ll find that the \( 3x^2 \) terms cancel, leaving you with: \[ -25x - 12 + 5x + 12 = -20x \] So, \( BC \) is \( 20x \) feet longer than \( AC \)! What a neat little mystery we solved here! One of the best ways to improve your skills in algebraic expressions is by practicing problems involving simplification and subtraction, as you now know how to derive dimensions through expressions. If you're tackling different variables, just remember to keep your terms organized to uncover the values quicker!

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