yuto left his house at 10 a.m. to go for a bike ride. by the time yutos sister riko left their house, yuto…

yuto left his house at 10 a.m. to go for a bike ride. by the time yutos sister riko left their house, yuto was already 5.25 miles along the path they both took. if yutos average speed was 0.25 miles per minute and rikos average speed was 0.35 miles per minute, over what time period in minutes, t, starting from when riko left the house, will riko be behind her brother? riko will be behind yuto when 0 ≤ t < minutes.

yuto left his house at 10 a.m. to go for a bike ride. by the time yutos sister riko left their house, yuto was already 5.25 miles along the path they both took. if yutos average speed was 0.25 miles per minute and rikos average speed was 0.35 miles per minute, over what time period in minutes, t, starting from when riko left the house, will riko be behind her brother? riko will be behind yuto when 0 ≤ t < minutes.

Answer

Answer:

52.5

Explanation:

Step1: Set up distance - equations

Let $d_Y$ be Yuto's distance and $d_R$ be Riko's distance. Yuto's initial head - start is 5.25 miles and his speed $v_Y = 0.25$ miles per minute, so $d_Y=5.25 + 0.25t$. Riko's speed $v_R = 0.35$ miles per minute, so $d_R = 0.35t$.

Step2: Set up inequality

We want to find when $d_R<d_Y$. Substitute the distance equations into the inequality: $0.35t<5.25 + 0.25t$.

Step3: Solve the inequality

Subtract $0.25t$ from both sides: $0.35t-0.25t<5.25$, which simplifies to $0.1t<5.25$. Then divide both sides by 0.1: $t < \frac{5.25}{0.1}=52.5$.