Answer:
The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.
Step-by-step explanation:
This is a problem of optimization.
We have to minimize the time it takes for the lifeguard to reach the child.
The time can be calculated by dividing the distance by the speed for each section.
The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.
Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:
Then, the time (speed divided by distance) is:
To optimize this function we have to derive and equal to zero:
![\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\](https://tex.z-dn.net/?f=%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%2B%5Cdfrac%7B1%7D%7B1.1%7D%28%5Cdfrac%7B1%7D%7B2%7D%29%5Cdfrac%7B2x-120%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%5C%5C%5C%5C%5C%5C%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%20%2B%5Cdfrac%7B1%7D%7B1.1%7D%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D0%5C%5C%5C%5C%5C%5C%20%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D%5Cdfrac%7B1.1%7D%7B4%7D%3D%5Cdfrac%7B2%7D%7B7%7D%5C%5C%5C%5C%5C%5C%20x-60%3D%5Cdfrac%7B2%7D%7B7%7D%5Csqrt%7Bx%5E2-120x%2B5200%7D%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B2%5E2%7D%7B7%5E2%7D%28x%5E2-120x%2B5200%29%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B4%7D%7B49%7D%5B%28x-60%29%5E2%2B40%5E2%5D%5C%5C%5C%5C%5C%5C%281-4%2F49%29%28x-60%29%5E2%3D4%2A40%5E2%2F49%3D6400%2F49%5C%5C%5C%5C%2845%2F49%29%28x-60%29%5E2%3D6400%2F49%5C%5C%5C%5C45%28x-60%29%5E2%3D6400%5C%5C%5C%5C)
As 
, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.
Answer:
3
x(x^2-3)
4x(x^2+5488)
Step-by-step explanation:
Answer:
Step-by-step explanation:
560=3s+b
b.
560=3×150+b
b=560-450=110
when s=149,b=560-3×149=560-447=113
s=2,b=560-3×2=560-6=554
s=1,b=560-3×1=557
s=0,b=560-3×o=560
so domain of b={110,113,116,...,554,557,560}
Answer: 2 2/5 is 4/5 of 3
Step-by-step explanation:
We can view that as what do we multiply 3 by to get 2 2/5 and call that value x
3x = 2 2/5
x = (2 2/5)/3
Let's turn 2 2/5 into an improper fraction... 2 2/5 = 12/5
So we have
x = (12/5)/3 = (12/5)(1/3) = 4/5
Thus 2 2/5 is 4/5 of 3
Verify: 3(4/5) = 12/5 = 2 2/5
I don't agree with his statement. The first part says that he was paid $9.20 for the first 40 hours so we multiply them (9.20 x 40) to get 368. Then it says that he was paid 1.5 times that rate for every hour after that. The rate is "$9.20 per hour" so you find 1.5 times $9.20. You multiply $9.20 by 1.5 to get $13.80 per hour. Now you have to find how much money he got after the initial 40 hours. It says he worked 42.25 hours, and we already figured out how much money he got the first 40 hours ($368). So for the remaining 2.25 hours, Mr.Evens is paid $13.80 an hour. So we multiply 2.25 by $13.80 to get $31.05. To find out the total amount of money accumulated, we add. $368 + $31.05 is $399.05 which means that Mr.Evens is not correct. Hope this helped :]
