Mathematics: Amotorboat traveled 35 km upstream on a river and then up an adjacent stream for 18 km, spending 8 hours on the entire trip. the speed of the river’s current was 1 km/hour slower than the speed of the stream’s current. find the speed of the current of the river, if the speed of the motorboat in still water is 10 km/hour.

Amotorboat traveled 35 km upstream on a river and

Ответ:

DogeDog6954

09.01.2022

Mathematics

3 mph

Step-by-step explanation:

Use distance (d) = rate (r) x time (t)

The rate is the rate of the boat in still water minus the current. The current is subtracted since you are going upstream.

$\begin {array}{l|c|c|c||l}&\underline{Time}&\underline{Rate}&\underline{Distance}&\underline{\qquad t\cdot r=d\qquad }\\River&x&10-c&35&x(10-c)=35\\Stream&8-x&10-(c+1)&18&(8-x)(9-c)=18\\\end{array}\\\\\\\text{Set each equation equal to x:}\\Stream:(8-x)(9-c)=18\qquad \qquad \quad River: x(10-c)=35\\.\qquad \qquad \qquad \quad \ 8-x=\dfrac{18}{9-c}\qquad \qquad \qquad \qquad \qquad x=\dfrac{35}{10-c}\\\\.\qquad \qquad \qquad \quad \ -x=\dfrac{18}{9-c}-8$

$.\qquad \qquad \qquad \quad \ x=\dfrac{-18}{9-c}+8\\\\\\\text{Since both equations are equal to x, we can set them equal to each other}\\\text{to solve for c}.\\\\\dfrac{35}{10-c}=\dfrac{-18}{9-c}+8\\\\\\\dfrac{35}{10-c}=\dfrac{-18}{9-c}+\dfrac{8(9-c)}{9-c}\\\\\\\dfrac{35}{10-c}=\dfrac{-18+72-8c}{9-c}\\\\\\\dfrac{35}{10-c}=\dfrac{54-8c}{9-c}\\\\\\\text{\underline{Cross multiply to solve for c:}}\\35(9-c)=(10-c)(54-8c)\\315-35c=540-80c-54c+8c^2\\315-35c=540-134c+8c^2$

$\text{\underline{Set the quadratic equal to zero and solve using quadratic formula:}}\\0=225-99c+8c^2\qquad \rightarrow \qquad a=8, b=-99, c=225\\\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\.\ =\dfrac{-(-99)\pm \sqrt{(-99)^2-4(8)(225)}}{2(8)}\\\\\\.\ =\dfrac{99\pm \sqrt{2601}}{16}\\\\\\.\ =\dfrac{99\pm 51}{16}\\\\\\.\ =\dfrac{99+ 51}{16}\quad or\quad \dfrac{99- 51}{16}\\\\\\.\ =\dfrac{150}{16}\quad or\quad \dfrac{48}{16}\\\\\\.\ =9.375\quad or\quad 3$

Since the rate of the Stream is 9 - c, the current cannot be 9.375 so that is an erroneous solution. Therefore, the current is 3

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