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Ymorist [56]
3 years ago
13

A racing shell, which is a boat a crew team rows, moves 20 km/h in still water. During practice, the team rows 36 km against the

current and 22 km with the current. Find the speed of the river current if the team rows a total of 3 hours.
Mathematics
2 answers:
saveliy_v [14]3 years ago
4 0

Answer:

2 km/h

Step-by-step explanation:

Let x represent the speed of the river current, t₁ be the time spent rowing against the current and t₂ be the time spent rowing with the current.

Since the speed of the still water = 20 km/h.

The speed of rowing against the current = (20 - x)

The speed of rowing with the current = (20 + x)

We know that velocity = distance / time. Hence time = distance / velocity

t₁ = 36 / (20 - x)

t₂ = 22 / (20 + x)

but t₁ + t₂ = 3 hours

Therefore:

3 = 36 / (20 - x) + 22 / (20 + x)

multiply through by 400 - x²

3(400 - x²) = 36(20 + x) + 22(20 - x)

1200 - 3x² = 720 + 36x + 440 - 22x

3x² + 14x -40 = 0

This gives x = -6.7 or x = 2

x cannot be negative therefore x = 2 km/h

The speed of the river current is 2 km/h

Ann [662]3 years ago
4 0

Answer:

2km/h is correct

Step-by-step explanation:

The other guy is absolutely correct

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Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?
likoan [24]

\mathrm{Use\:the\:rational\:root\:theorem}

a_0=15,\:\quad a_n=2

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5

=\left(x+3\right)\left(2x^2-11x+5\right)

Factor: 2x^2-11x+5

2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)

=x\left(2x-1\right)-5\left(2x-1\right)

2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)

\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:}

thus zeros of f(x) is

x=-3,\:x=5,\:x=\frac{1}{2}

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Can you help me please.
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Answer:

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