Aturtle and a snail agreed to have a race. being kind, the turtle gave the snail a head start. the snail took off at a blinding 3 feet per minute. 50 minutes later, the turtle went after him at a blistering 9 feet per minute s(m) = 150 + 3m models the snail's distance after the turtle started, where m is the minutes. t (m) = 9m models the turtle's distance. in how many minutes will the turtle catch up to the snail? a. 25 minutes b. 35 minutes c. 30 minutes d. 45 minutes

In 25 minutes turtle will catch up to the snail.

Step-by-step explanation:

we have given,

Snail's speed = 3 feet per minute

Snail's had a head start , so after 50 minutes Snail's distance = 150

Now after 50minute we can model snail's distance equation for m minutes as:

S(m) = 150 + 3m

next, turtle went after him after 50 minutes at a speed of 9 feet per minute, and the distance turtle can cover in m minutes will be:

Turtle's distance, T(m) = 9m

We need to calculate the minutes when turtle catch up to the snail.

i.e distance covered by snails and turtle will equal at the point where they meet ,

S(m) = T(m)  

or 150 + 3m = 9m

or 150 = 9m - 3m

or 150 = 6m

or m = minutes

So, In 25 minutes turtle will catch up to the snail.

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.