An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is 15 mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of 6 mph and swims at a rate of 2.5 mph. How far should the visitor run before swimming to minimize the time it takes to reach the island

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim =

The total time, 't', is given as follows;

The minimum value of 't' is found by differentiating with an online tool, as follows;

At the maximum/minimum point, we have;

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

Therefore, the distance to run, x ≈ 14.96 mile

75

step-by-step explanation:

you do x-29=46 (the number of cantaloupes sold)

in order to find x, you have to find what minus 29 equals 46, meaning you'd have to add 29 to 46.

29+46=75

if you replace x with 75, the equation looks like:

75-29=46

x=75

the answer is 40.

proof: