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elixir [45]
3 years ago
6

A boat travels 2km upstream and 2km downstream. The total time of the trip is 1 hour. The speed of the stream is 2km/h. What is

the speed of the boat in still water?
Mathematics
1 answer:
scoray [572]3 years ago
7 0
2 = ( v + 2 ) × t1 ;
2 = ( v - 2 ) × t2 ;
t1 + t2 = 1 ;
where v is the speed of the boat in still water ; t1 is the time upstream ; t2 is the tome downstream ;
Then, v + 2 = 2 / t1 and v - 2 = 2 / t2 ;
v = 2 / t1   -  2  and v = 2 / t2  +  2 ;
2 / t1 - 2 = 2 / t2 + 2 ;
2 / t1 - 2 / t2 = 4 ;
1 / t1 - 1 / t2 = 2 ;
t2 - t1 = 2t1t2 ;
But, t2 = 1 - t1 ;
1 - t1 - t1 = 2t1( 1 - t1 ) ;
1 - 2t1 = 2t1 - 2t1^2 ;
2t1^2 - 4t1 + 1 = 0 ; quadratic equation ;
solving this equation , 
t1 ≈ 0.29 hour ;
v ≈ 2 / 0.29 - 2;
v ≈ 4.89 km/h ;

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Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

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The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345

3.h(x)=\frac{3x-5}{x^2-5x+7}

x^2-5x+7=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43

2.g(x)=\frac{x-17}{x^2+75}

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12

4.i(x)=\frac{x^2-9}{x-9}

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is  not continuous on the set of real numbers.

5.j(x)=\frac{4x^2-7x-65}{x^2+10}

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0

6.k(x)=\frac{x+1}{x^2+x+29}

x^2+x+29=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102

7.l(x)=\frac{5x-1}{x^2-9x+8}

x^2-9x+8=0

x^2-8x-x+8=0

x(x-8)-1(x-8)=0

(x-8)(x-1)=0

x=8,1

The function is not defined for x=8 and x=1

Hence, function l is not  defined for all real values.

8.m(x)=\frac{x^2+5x-24}{x^2+11}

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722

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Answer:

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raketka [301]
Given,
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Answer:

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Step-by-step explanation:

Parallel lines have the same slope.

Changing the constant in a linear equation like this only changes the y-intercept. It has no effect on the slope of the line. So, we can change the constant from 4 to 0 and we will have a line with the same slope, parallel to the original, but with a different y-intercept.

The "standard form" of the equation of a line has the leading coefficient positive. We can make that be the case by using the multiplication property of equality, multiplying both sides of the equation by -1.

Parallel line:

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