Solved on Sep 24, 2023

Latoya has two exercise routines. Routine #1: 22 calories walking, $15.5$ calories/min running. Routine #2: 40 calories walking, $13.25$ calories/min running. Find time $t$ (min) running where Routine #1 burns at most as many calories as Routine #2.

STEP 1

Assumptions1. In Routine #1, Latoya burns22 calories walking and then burns15.5 calories per minute running. . In Routine #, Latoya burns40 calories walking and then burns13.25 calories per minute running.
3. We use $t$ to represent the number of minutes spent running.
4. We want to find for what amounts of time spent running will Routine #1 burn at most as many calories as Routine #.

STEP 2

First, we need to set up an inequality to represent the problem. The total calories burned in each routine is the sum of the calories burned walking and the calories burned running. We want the total calories burned in Routine #1 to be at most (less than or equal to) the total calories burned in Routine #2.
$22 +15.5t \leq40 +13.25t$

STEP 3

Next, we need to isolate the variable $t$ on one side of the inequality. We can start by subtracting $13.25t$ from both sides of the inequality.
$22 +15.5t -13.25t \leq40 +13.25t -13.25t$

STEP 4

implify the inequality.
$22 +2.25t \leq40$

STEP 5

Next, subtract22 from both sides of the inequality to isolate the term with $t$ on one side of the inequality.
$22 +2.25t -22 \leq40 -22$

STEP 6

implify the inequality.
$2.25t \leq18$

STEP 7

Finally, divide both sides of the inequality by2.25 to solve for $t$ .
$\frac{2.25t}{2.25} \leq \frac{18}{2.25}$

STEP 8

implify the inequality to find the solution for $t$ .
$t \leq8$ So, for Latoya to burn at most as many calories in Routine #1 as in Routine #2, she should run for at most8 minutes.

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Latoya has two exercise routines. Routine #1: 22 calories walking, $15.5$ calories/min running. Routine #2: 40 calories walking, $13.25$ calories/min running. Find time $t$ (min) running where Routine #1 burns at most as many calories as Routine #2.

STEP 1

STEP 2

STEP 3

Next, we need to isolate the variable $t$ on one side of the inequality. We can start by subtracting $13.25t$ from both sides of the inequality.
$22 +15.5t -13.25t \leq40 +13.25t -13.25t$

STEP 4

implify the inequality.
$22 +2.25t \leq40$

STEP 5

Next, subtract22 from both sides of the inequality to isolate the term with $t$ on one side of the inequality.
$22 +2.25t -22 \leq40 -22$

STEP 6

implify the inequality.
$2.25t \leq18$

STEP 7

Finally, divide both sides of the inequality by2.25 to solve for $t$ .
$\frac{2.25t}{2.25} \leq \frac{18}{2.25}$

STEP 8

implify the inequality to find the solution for $t$ .
$t \leq8$ So, for Latoya to burn at most as many calories in Routine #1 as in Routine #2, she should run for at most8 minutes.

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Latoya has two exercise routines. Routine #1: 22 calories walking, 15.515.515.5 calories/min running. Routine #2: 40 calories walking, 13.2513.2513.25 calories/min running. Find time ttt (min) running where Routine #1 burns at most as many calories as Routine #2.

STEP 1

STEP 2

STEP 3

Next, we need to isolate the variable ttt on one side of the inequality. We can start by subtracting 13.25t13.25t13.25t from both sides of the inequality.22+15.5t−13.25t≤40+13.25t−13.25t22 +15.5t -13.25t \leq40 +13.25t -13.25t22+15.5t−13.25t≤40+13.25t−13.25t

STEP 4

implify the inequality.22+2.25t≤4022 +2.25t \leq4022+2.25t≤40

STEP 5

Next, subtract22 from both sides of the inequality to isolate the term with ttt on one side of the inequality.22+2.25t−22≤40−2222 +2.25t -22 \leq40 -2222+2.25t−22≤40−22

STEP 6

implify the inequality.2.25t≤182.25t \leq182.25t≤18

STEP 7

Finally, divide both sides of the inequality by2.25 to solve for ttt.2.25t2.25≤182.25\frac{2.25t}{2.25} \leq \frac{18}{2.25}2.252.25t​≤2.2518​

STEP 8

implify the inequality to find the solution for ttt.t≤8t \leq8t≤8So, for Latoya to burn at most as many calories in Routine #1 as in Routine #2, she should run for at most8 minutes.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Latoya has two exercise routines. Routine #1: 22 calories walking, $15.5$ calories/min running. Routine #2: 40 calories walking, $13.25$ calories/min running. Find time $t$ (min) running where Routine #1 burns at most as many calories as Routine #2.

Next, we need to isolate the variable $t$ on one side of the inequality. We can start by subtracting $13.25t$ from both sides of the inequality.
$22 +15.5t -13.25t \leq40 +13.25t -13.25t$

implify the inequality.
$22 +2.25t \leq40$

Next, subtract22 from both sides of the inequality to isolate the term with $t$ on one side of the inequality.
$22 +2.25t -22 \leq40 -22$

implify the inequality.
$2.25t \leq18$

Finally, divide both sides of the inequality by2.25 to solve for $t$ .
$\frac{2.25t}{2.25} \leq \frac{18}{2.25}$

implify the inequality to find the solution for $t$ .
$t \leq8$ So, for Latoya to burn at most as many calories in Routine #1 as in Routine #2, she should run for at most8 minutes.