Mia climbs 2/5 of the height of the rock wall. Lee climbs 3/8 of Mia's distance. What fraction of the wall did Lee climb?
1 answer:
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Part A:
Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.
Let the length of the other side be x, then
Thus, if the length of one of the side is x, the length of the other side is 8 - L.
Hence, the area of the rectangle in terms of L is given by
Part B:
To find the domain of A
Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.
Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.
Hence the domain of the area are the set of values between 0 and 8 not inclusive.
Therefore,
Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.
Let the length of the other side be x, then
Thus, if the length of one of the side is x, the length of the other side is 8 - L.
Hence, the area of the rectangle in terms of L is given by
Part B:
To find the domain of A
Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.
Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.
Hence the domain of the area are the set of values between 0 and 8 not inclusive.
Therefore,
Answer:
Santos should run approximately 36.145 meters along the shore
Step-by-step explanation:
The given parameters are;
The distance along the shore of the child = 50 meters
The distance from the shore of the child = 60 meters
The rate at which Santos can run = 4 m/s
The rate he can swim = 0.9 m/s
Let 'x' represent the distance he runs along the shore
We have;
The time he spends running on the shore, t₁ = x/4
The time he spends swimming, t₂ = (√(60² + (50 - x)²)/0.9
The total time, T = t₁ + t₂ = x/4 + (√(60² + (50 - x)²)/0.9
To find the maximum, we have;
dT/dx = 0 = d(x/4 + (√(60² + (50 - x)²)/0.9)/dx = 1/4 - (50 - x)/(0.9·(√(60² + (50 - x)²) = 0
1/4 - (50 - x)/(0.9·(√(60² + (50 - x)²) = 0
Simplifying using a graphing calculator gives;
1519·x² - 151900·x + 3505900 = 0
x = (151900 ± √((-151900)² - 4×1519×3505900))/(2 × 1519)
x ≈ 63.86 m or x ≈ 36.145 m
We note that the distance from a point x = 63.83 meters and 36.145 meters from where Santos spots the girl to the location of the girl are the same
Therefore, Santos should run approximately 36.145 meters along the shore before jumping into to the water in order to save the child
