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ANEK [815]
3 years ago
14

A slow boat to china travels at a constant speed of 17.25 miles per hour for 200 hours. how far was the voyage?​

Mathematics
2 answers:
AleksAgata [21]3 years ago
7 0

Answer:

\boxed {\boxed {\sf 3,450 \ miles}}

Step-by-step explanation:

Distance can be found by multiply the speed by the time.

d=st

Where <em>d</em> is the distance, <em>s</em> is the speed, and <em>t</em> is the time.

1. Define Variables

The speed is 17.25 miles per hour and the time is 200 hours.

s= 17.25 \ mi/hr

t= 200 \ hr

2. Calculate Speed

Substitute the values into the formula.

d=17.25 \ mi/hr * 200 \ hr

Multiply. The hours (hr) will cancel each other out and leave miles as the units.

d=17.25 \ mi *200\\

d=3450 \ mi

The voyage was <u>3,450 miles</u>

tensa zangetsu [6.8K]3 years ago
4 0

Answer:

3450 miles

Step-by-step explanation:

Multiply the number 17.25 by 200 since the boat is traveling for 200 hours and each hour is 17.25 miles to get 3450 miles which is your answer.

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Critical value f(1)=2.

Minimum at (1,2), function is decreasing for 0 and increasing for x>1.

\left(3,\dfrac{4}{\sqrt{3}}\right) is point of inflection.

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Step-by-step explanation:

1. Find the domain of the function f(x):

\left\{\begin{array}{l}x\ge 0\\x\neq 0\end{array}\right.\Rightarrow x>0.

2. Find the derivative f'(x):

f'(x)=\dfrac{(x+1)'\cdot \sqrt{x}-(x+1)\cdot (\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}=\dfrac{2x-x-1}{2x\sqrt{x}}=\dfrac{x-1}{2x^{\frac{3}{2}}}.

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2. For 0 the derivative f'(x)<0, then the function is decreasing. For x>1, the derivative f'(x)>0, then the function is increasing. This means that point x=1 is point of minimum.

3. Find f''(x):

f''(x)=\dfrac{(x-1)'\cdot 2x^{\frac{3}{2}}-(x-1)\cdot (2x^{\frac{3}{2}})'}{(2x^{\frac{3}{2}})^2}=

=\dfrac{2x^{\frac{3}{2}}-2(x-1)\cdot \frac{3}{2}x^{\frac{1}{2}}}{4x^3}=\dfrac{2x^{\frac{3}{2}}-2\cdot\frac{3}{2}x^{\frac{3}{2}}+ 2\cdot\frac{3}{2}x^{\frac{1}{2}}}{4x^3}=

=\dfrac{-x+3}{4x^{\frac{5}{2}}}.

When f''(x)=0, x=3 and f(3)=\dfrac{3+1}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}.

When 0<x<3, f''(x)>0 - function is concave upwards and when x>3, f''(x)>0 - function is concave downwards.

Point \left(3,\dfrac{4}{\sqrt{3}}\right) is point of inflection.

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