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.
After the 15% discount, Maria pays 85% of the cost. That cost is the sum of the fixed cost and the variable (per visit) cost.
... (updated cost) = 0.85×($150 +$30×a)
... (updated cost) = 127.50 +25.50a . . . . in dollars
Answer:
- D. A translation 1 unit to the right followed by a 270-degree counterclockwise rotation about the origin
Step-by-step explanation:
<em>See the picture for better visual</em>
Take segments ST and S'T'. If we extend them they will intersect at right angle.
It is the indication that the rotation is 90° or 270° but not 180°, when the corresponding segments come parallel.
The QRST is in the quadrant IV and Q'R'S'T' is in the quadrant III, which mean the rotation is 90° clockwise or 270° counterclockwise.
<u>This rotation rule is:</u>
We also see the points S and T have x-coordinate of 5 but their images have y-coordinates of -6. It means the translation to the right by 1 unit was the step before rotation.
<u>We now can conclude the correct choice is D:</u>
- A translation 1 unit to the right followed by a 270-degree counterclockwise rotation about the origin
Answer:
The answer I think is correct is B
Step-by-step explanation:
I hope that's corrected?
Answer:
11/50
Step-by-step explanation:
The frequency numbers for landing on 1, 2, and 3 are:
25, 53, 62.
We add them up to get: 25 + 53 + 62 = 140
Since 250 spins were made, and 140 of them landed on 1, 2, or 3, then
250 - 140 = 110,
so 110 landed on 4 or 5.
We are told the numbers of spins landing on 4 and 5 are equal, so
110/2 = 55,
so the spinner landed 55 times on 4 and 55 times on 5.
relative frequency = 55/250 = 11/50
