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SIZIF [17.4K]
3 years ago
11

Miguel and his brother Ario are both standing 3 meters from one side of a 25-meter pool when they decide to race. Miguel offers

Ario a head start. Miguel says he will start when the ratio of Ario's completed meters to Ario's remaining meters is 1:4. How many meters will Ario swim before Miguel starts the race? Round to the nearest tenth. meters
Mathematics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

Miguel start to swim when Ario has traveled 4.4 m.

Step-by-step explanation:

The total lenght of the pool is 25.

Since both of them standing 3 m from one side of the pool, then, the total distance both need to cover is 25-3 m= 22 m.

Assume that the distance traveled by Ario before Miguel start to swin is x. That means, the remaining distance to Ario (to cover the total size of the pool) is 22-x.

Then, in acoordance to the problem, the ratio between these two distances must be equal to 1/4.

That is,

\frac{x}{22-x}=\frac{1}{4} .

So, we need to obtain x from this equation. We must note that x\neq 22, otherwise we have a zero in the denominator.

So, we rearrange the equation,

4x=22-x (Multiplying both sides of equation by (4(22-x)).

Then, 5x=22.

Therefore, x=4.4 m.

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Step-by-step explanation:

Given

f(x) = \sin x\\

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Required

Find the Taylor series

The Taylor series of a function is defines as:

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

We have:

c = \frac{3\pi}{4}

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f(c) = \sin(c)

f(c) = \sin(\frac{3\pi}{4})

This gives:

f(c) = \frac{1}{\sqrt 2}

We have:

f(c) = \sin(\frac{3\pi}{4})

Differentiate

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This gives:

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We have:

f'(c) = \cos(\frac{3\pi}{4})

Differentiate

f"(c) = -\sin(\frac{3\pi}{4})

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f"(c) = -\frac{1}{\sqrt 2}

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f"(c) = -\sin(\frac{3\pi}{4})

Differentiate

f"'(c) = -\cos(\frac{3\pi}{4})

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f"'(c) = - * -\frac{1}{\sqrt 2}

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f'(c) = -\frac{1}{\sqrt 2}

f"(c) = -\frac{1}{\sqrt 2}

f"'(c) = \frac{1}{\sqrt 2}

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

becomes

f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n

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Hence:

\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

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