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vovikov84 [41]
3 years ago
5

HEEEELPPPPPPPPPPPPPP!!!!!!!!!!

Mathematics
1 answer:
Marianna [84]3 years ago
4 0
B- use Desmos it helps a lot!
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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 i
Elanso [62]

Answer:

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 = \sqrt{(15-x)^2+1^2}

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

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

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

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

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

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

6 0
3 years ago
What does Fish equal?<br> A Nothing <br> B 2+2<br> C More Fish
zhuklara [117]

Answer:

A nothing

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Solve the equation 5(a – 7) + 11 = 2a – 3(a + 4) for a
Lady bird [3.3K]
The answer is a equals 6.
6 0
3 years ago
Read 2 more answers
-3x²+9x-6 find the terms, constant, Factors, and Coefficients.
Anastasy [175]

Answer:

WAIT A MINUTE MR. POSTMAN

Step-by-step explanation:

7 0
3 years ago
Please help with this geometry question
Andrew [12]

It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

<h3>How to prove a Line Segment?</h3>

We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.

Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.

In ΔPNM, ∠N = 90°

∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)

∠P + ∠M = 90°

Clearly, ∠M is an acute angle.

Thus; ∠M < ∠N

PN < PM (The side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Read more about Line segment at; brainly.com/question/2437195

#SPJ1

7 0
2 years ago
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