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Determining the Difference in Distance

Points A, B, and C form a triangle. The distance between point A and point B is 15 yards. The distance between point B and point C is 25 yards. Pete walks directly from point A to point C, without passing through point B.

1. What is the direct distance from A to C?
2. How far would Pete walk if he went from A to B to C?
3. The direct distance from A to C is more than 10 yards. The inequality [tex][tex]$\ \textless \ $[/tex][/tex] represents the distance, w, that Pete might save by taking the direct path.

Sagot :

GinnyAnswer
Sure, let's break down the problem and formulate a complete solution step-by-step:

1. We are given a triangle with points A, B, and C.
2. The distance from point A to point B (AB) is 15 yards.
3. The distance from point B to point C (BC) is 25 yards.
4. Pete walks directly from point A to point C, which is not passing through point B.

To find the distances:
- The total walking distance if Pete goes from A to B to C (i.e., along the path A ā†’ B ā†’ C) is:
[tex]\[ AB + BC = 15 \text{ yards} + 25 \text{ yards} = 40 \text{ yards} \][/tex]

- The direct distance from A to C (AC) should be calculated, and we are told that the direct distance is greater than 10 yards. However, there is no information provided about the exact direct distance from A to C. In the absence of additional information, let's use the most likely geometric interpretation:

Since [tex]\( AC \)[/tex] should be straight,
This calculation's result (AC = 40 yards) is the total path A ā†’ B ā†’ C accelerates error potentially without the actual direct distance known. For practical framework, the correct direct distance needs explicit information or method for direct linear potential structure configuring.

To determine the inequality that represents the distance Pete might save by taking the direct path:
- If Pete walks the path A ā†’ B ā†’ C, he walks a total of 40 yards.
- Pete saves by not walking via B.

Since [tex]\( AC \neq AB + BC = 40 \text{ yards accurately unless expressed direct path }, the saved distance ( \( w \)[/tex] ) also indirectly spans within ranges which leads:

- Given [tex]\( w \)[/tex] as yard's inequality positioning:
[tex]\[ \text{Saved distance lower bound } > 10 \text{ yards} \][/tex]
As bounds, evaluated distance deviation from \( >10 \text{ and till \( upper bound = 30 mis-hly without accurate AC specific defining, a ā„¹ analytical rationale mismatch debugging back for linear AC.} )

Thus, resulting:
- \( w :\text{\direct while present solves ( \( AC precise measure unecho!, expansive hypothetical direct arrange eats ) ā‰„ 10 upwards while potential } 30. \text{ (actualizing AC not linear express).}

In conclusion:

1. Pete saves more than 10 yards taking the direct distance ( \( \text >10\footin potential no actual given} )
2. Configuring \( ( though standard scorching spans} \( 30 wrong extended max till proven specific)

Answer for result :
:
AC: \text {missing linear path appropriates correcting '}
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