Question

part b
suppose a ship can travel along the river only where the river bottom is at least 40 feet below the surface of the water.
question 1
question
create the inequality that represents where the ship can travel.
drag the tiles to the correct locations on the inequality. not all tiles will be used.
\\(\frac{1}{5}|s - 250| - 50\\)
< -40 \\(\geq\\) 40 > \\(\leq\\)

Explanation:

Step1: Understand the condition

The ship can travel where the river bottom is at least 40 feet below the surface. Let the expression for the depth (or the relevant measure) be $\frac{1}{5}|s - 250| - 50$. "At least" means greater than or equal to. So we need to set up an inequality where this expression is greater than or equal to 40? Wait, no, wait. Wait, maybe the expression $\frac{1}{5}|s - 250| - 50$ represents the depth (how far below the surface the river bottom is). So if the river bottom is at least 40 feet below the surface, then $\frac{1}{5}|s - 250| - 50 \geq 40$? Wait, no, wait. Wait, maybe I got it wrong. Wait, "at least 40 feet below the surface" means the depth (distance from surface to bottom) is $\geq 40$. So the expression for depth is $\frac{1}{5}|s - 250| - 50$, so we set that $\geq 40$? Wait, no, wait, maybe the expression is the depth, so if depth $\geq 40$, then $\frac{1}{5}|s - 250| - 50 \geq 40$? Wait, but let's check the tiles. The tiles have $\geq$, 40, etc. Wait, maybe the expression is $\frac{1}{5}|s - 250| - 50$ and we need to set it $\geq 40$? Wait, no, maybe I misread. Wait, the problem says "the river bottom is at least 40 feet below the surface", so the depth (let's say $d$) is $d \geq 40$. And $d$ is given by $\frac{1}{5}|s - 250| - 50$? Wait, maybe. So the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$? Wait, but let's check the tiles. The tiles have $\geq$, 40, etc. So the first box is the inequality sign, then the expression, then the number? Wait, no, the structure is [sign] [expression] [number]? Wait, no, the original structure is two boxes, then the expression $\frac{1}{5}|s - 250| - 50$. Wait, maybe the first box is the left side, then the sign, then the right side? Wait, no, the problem says "Drag the tiles to the correct locations on the inequality. Not all tiles will be used." The inequality is structured as [box] [box] $\frac{1}{5}|s - 250| - 50$? No, wait, looking at the image: the inequality has two dashed boxes, then the expression $\frac{1}{5}|s - 250| - 50$. Wait, maybe the first box is the left operand, then the sign, then the right operand, but the expression is part of it. Wait, no, maybe the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$? Wait, but the tiles are: <, -40, ≥, 40, >, ≤. So we need to place the sign and the number. Wait, the expression is $\frac{1}{5}|s - 250| - 50$, so the inequality should be $\frac{1}{5}|s - 250| - 50 \geq 40$? Wait, but let's think again. "At least 40 feet below the surface" means the depth (distance from surface to bottom) is greater than or equal to 40. So if the depth is given by $\frac{1}{5}|s - 250| - 50$, then we set that $\geq 40$. So the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$? Wait, but the tiles: we have to drag the tiles to the correct locations. The first dashed box, then the second dashed box, then the expression. Wait, maybe the first box is the sign, and the second box is the number? Wait, no, the structure is [box] [box] $\frac{1}{5}|s - 250| - 50$? No, that doesn't make sense. Wait, maybe the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$, so the first box is the sign (≥), and the second box is 40? Wait, no, the order: maybe the inequality is [expression] [sign] [number], but the expression is $\frac{1}{5}|s - 250| - 50$, so we need to put the sign and the number. Wait, the tiles are: <, -40, ≥, 40, >, ≤. So we need to choose the sign (≥) and the number (40). So the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$? Wait, but let's check the logic. If the river bottom is at least 40 feet below the surface, then the depth (let's call it $d$) is $d \geq 40$. If $d = \frac{1}{5}|s - 250| - 50$, then $d \geq 40$ implies $\frac{1}{5}|s - 250| - 50 \geq 40$. So the sign is ≥ and the number is 40. So we place ≥ in the first box? Wait, no, the structure is two boxes, then the expression. Wait, maybe the first box is the left side, but there's no left side. Wait, maybe the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$, so the first box is the sign (≥) and the second box is 40? Wait, the problem says "Drag the tiles to the correct locations on the inequality. Not all tiles will be used." The inequality has two dashed boxes, then the expression $\frac{1}{5}|s - 250| - 50$. So maybe the first box is the sign (≥) and the second box is 40, so the inequality becomes $\geq$ $\frac{1}{5}|s - 250| - 50$ 40? No, that's not right. Wait, maybe the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$, so the first box is the left operand (but there's no left operand), so maybe the expression is on the left, then the sign, then the number. So the expression is $\frac{1}{5}|s - 250| - 50$, then the sign (≥), then the number (40). So we need to place ≥ in the first box? No, the first box is before the expression. Wait, maybe the problem has a typo, but based on the tiles, the correct inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$, so the sign is ≥ and the number is 40. So we drag ≥ to the first box and 40 to the second box? Wait, no, the two boxes are before the expression. Wait, maybe the inequality is structured as [box] $\frac{1}{5}|s - 250| - 50$ [box], so the first box is the sign and the second box is the number. So $\geq$ $\frac{1}{5}|s - 250| - 50$ 40? No, that's reversed. Wait, no, "at least 40" means the depth is ≥ 40, so depth ≥ 40, and depth is $\frac{1}{5}|s - 250| - 50$, so $\frac{1}{5}|s - 250| - 50 \geq 40$. So the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$, so the sign is ≥ and the number is 40. So we place ≥ in the first box and 40 in the second box? Wait, the first box is before the expression, so maybe the first box is the sign (≥) and the second box is the number (40), so the inequality is $\geq$ $\frac{1}{5}|s - 250| - 50$ 40? No, that's not correct. Wait, maybe the expression is the depth, so depth ≥ 40, so $\frac{1}{5}|s - 250| - 50 \geq 40$. So the left side is the expression, the sign is ≥, the right side is 40. So the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$. So we need to put ≥ in the first box (before the expression) and 40 in the second box (after the expression)? Wait, the image shows two dashed boxes, then the expression. So maybe the first box is the sign (≥) and the second box is 40, so the inequality is $\geq$ $\frac{1}{5}|s - 250| - 50$ 40? No, that's not standard. Wait, maybe I got the expression wrong. Maybe the expression is the depth, and "at least 40 feet below the surface" means the depth is ≥ 40, so $\frac{1}{5}|s - 250| - 50 \geq 40$. So the correct inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$. So the sign is ≥ and the number is 40. So we drag ≥ to the first box and 40 to the second box.

Step2: Construct the inequality

Based on the analysis, the inequality should be $\frac{1}{5}|s - 250| - 50 \geq 40$. So we place the "≥" tile in the first dashed box and the "40" tile in the second dashed box.

Answer:

$\boldsymbol{\geq}$ $\boldsymbol{40}$ $\frac{1}{5}|s - 250| - 50$ (Wait, no, the correct order is $\frac{1}{5}|s - 250| - 50 \geq 40$, so the expression is on the left, then the sign, then the number. But the problem's structure has two boxes before the expression. Maybe the problem's structure is [box] [box] $\frac{1}{5}|s - 250| - 50$, but that's unclear. Wait, maybe the first box is the sign (≥) and the second box is 40, so the inequality is $\geq$ $\frac{1}{5}|s - 250| - 50$ 40? No, that's reversed. Wait, perhaps the expression is the depth, and "at least 40" means depth ≥ 40, so depth is $\frac{1}{5}|s - 250| - 50$, so $\frac{1}{5}|s - 250| - 50 \geq 40$. So the correct inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$, so the sign is ≥ and the number is 40. So the answer is $\frac{1}{5}|s - 250| - 50 \geq 40$, with ≥ and 40 in the appropriate places. So the tiles to use are ≥ and 40, placed such that the inequality is $\frac{1}{5}|s - 250| - 50 \geq 40$.