Respuesta :
[tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
downstream&36&20+r&t\\
upstream&22&20-r&3-t
\end{array}[/tex]
as you know, going with the current (downstream), the speed rate of "r" is added to boat's still water rate, and upstream is subtracted
so.. anyhow, with it, it went 36km, against it it went 22km
hmmm the time it took was 3hrs total
so, if it took "t" time going downstream, then going upstream took the difference, 3 - t
so recalling your d = rt
[tex]\bf \begin{cases} 36=(20+r)t\implies \frac{36}{20+r}=t\\\\ 22=(20-r)(3-t)\\ ----------\\ 22=(20-r)\left( 3-\frac{36}{20+r}\right) \end{cases}\\\\ -----------------------------\\\\ 22=(20-r)\left( \cfrac{60+3r-36}{20+r}\right)\implies 22=(20-r)\left( \cfrac{24+3r}{20+r}\right) \\\\\\ 22(20+r)=(20-r)(24+3r) \\\\\\ 440+22r=480+60r-24r-3r^2 \\\\\\ 3r^2-14r-40=0[/tex]
now.. that one doesn't have integer values...so let's run it through the quadratic formula
[tex]\bf r=\cfrac{14\pm\sqrt{14^2+4(3)(-40)}}{6}\implies r=\cfrac{14\pm\sqrt{196+480}}{6} \\\\\\ r=\cfrac{14\pm\sqrt{676}}{6}\implies r=\cfrac{14\pm 26}{6}\implies r=\cfrac{7\pm 13}{3} \\\\\\ r= \begin{cases} \frac{20}{3}\\\\ -2\leftarrow \textit{not valid for this rate} \end{cases} \\\\\\ thus \qquad r=\cfrac{20}{3}[/tex]
as you know, going with the current (downstream), the speed rate of "r" is added to boat's still water rate, and upstream is subtracted
so.. anyhow, with it, it went 36km, against it it went 22km
hmmm the time it took was 3hrs total
so, if it took "t" time going downstream, then going upstream took the difference, 3 - t
so recalling your d = rt
[tex]\bf \begin{cases} 36=(20+r)t\implies \frac{36}{20+r}=t\\\\ 22=(20-r)(3-t)\\ ----------\\ 22=(20-r)\left( 3-\frac{36}{20+r}\right) \end{cases}\\\\ -----------------------------\\\\ 22=(20-r)\left( \cfrac{60+3r-36}{20+r}\right)\implies 22=(20-r)\left( \cfrac{24+3r}{20+r}\right) \\\\\\ 22(20+r)=(20-r)(24+3r) \\\\\\ 440+22r=480+60r-24r-3r^2 \\\\\\ 3r^2-14r-40=0[/tex]
now.. that one doesn't have integer values...so let's run it through the quadratic formula
[tex]\bf r=\cfrac{14\pm\sqrt{14^2+4(3)(-40)}}{6}\implies r=\cfrac{14\pm\sqrt{196+480}}{6} \\\\\\ r=\cfrac{14\pm\sqrt{676}}{6}\implies r=\cfrac{14\pm 26}{6}\implies r=\cfrac{7\pm 13}{3} \\\\\\ r= \begin{cases} \frac{20}{3}\\\\ -2\leftarrow \textit{not valid for this rate} \end{cases} \\\\\\ thus \qquad r=\cfrac{20}{3}[/tex]
The speed of the current of the river can be calculated by quadratic equation which is 33.89 approx.
What is roots of a quadratic equation ?
The roots of a quadratic equation are the values of the variable that satisfy the equation. They are also known as the "solutions" or "zeros" of the quadratic equation
For example, the roots of the quadratic equation x² - 7x + 10 = 0 are x = 2 and x = 5 x = (-b ± √ (b² - 4ac) )/2aThis is known as the quadratic formula and it can be used to find any type of roots of a quadratic equation.
According to the question Speed of boat in still water = 20 km/hour
Let speed of current of the river = x km/hour
Therefore, total speed of boat against the current = speed of current of the river - Speed of boat in still water = x - 20 Total speed of boat with the current = speed of current of the river + Speed of boat in still water = x + 20
Now, has traveled 36 km against the current and 22 km with the current, having spent 3 hours on the entire trip. i.e[tex]\frac{36}{x-20} + \frac{22}{x + 20} = 3[/tex] by solving the equation we will get [tex]x \frac{36}{x-20} + \frac{22}{x + 20}[/tex] = [tex]3\frac{36(x + 20) + 22(x-20)}{x^{2} -400} = 3\frac{36x + 720 + 22x-440}{x^{2} -400} = 3\frac{58x + 280}{x^{2} -400} = 358x + 280 = 3x^{2} -1200 0 = 3x^{2} - 58x - 1480[/tex]
Using roots of a quadratic equation x = (-b ± √ (b² - 4ac) )/2a substituting the values in formula[tex]x = \frac{58 + \sqrt{(-58)^{2} + 4*3* 1480 } }{2*3} x = 33.89[/tex] (approx.)
or [tex]x = \frac{58 - \sqrt{(-58)^{2} + 4*3* 1480 } }{2*3}[/tex] not possible as square root is in negative
Hence, the speed of the current of the river is 33.89 approx.
To know more about roots of a quadratic equation here:
https://brainly.com/question/26926523
#SPJ2
