Yuto left his house at 10 a. m. to go for a bike ride. by the time yuto’s sister riko left their house, yuto was already 5.25 miles along the path they both took. if yuto’s average speed was 0.25 miles per minute and riko’s 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.

-- When Riko left the house, Yuto was 5.25 miles ahead of her.

-- From the time she left the house, she gained (0.35 - 0.25) = 0.1 mile on Yuto

every minute ... closing the gap at the rate of 0.1 mile per minute.

-- If she can keep it up, she'll close the gap in (5.25 / 0.1) = 52.5 minutes.

-- Riko is behind her brother when  0 ≤ t < 52.5 minutes .

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

more detail

step-by-step explanation: