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Natalija [7]
3 years ago
8

Roberto rowed 20 miles downstream in 2.5 hours. The trip back, however, took him 5 hours. Find the rate that Roberto rows in sti

ll water and the rate of the current. If x is Roberto's rowing speed, and y is the speed of the river current, construct one equation modeling his downstream trip and another modeling his upstream trip using x and y .
The downstream equation is ____________

The upstream equation is ___________
Mathematics
1 answer:
almond37 [142]3 years ago
3 0

Answer:

<em>Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour</em>

Step-by-step explanation:

<u>Relative Speed</u>

When a body is moving at a constant speed v, the distance traveled in a time t is:

d=v.t

When Roberto rows downstream, his speed in still water is added to the speed of the water, making it easier to travel the required distance.

When Roberto rows upstream, his speed in still water is affected by the speed of the water, both are subtracted and the required distance is covered in more time.

Let's call

x = Roberto's rowing speed in still water

y = Speed of the river current

The speed when rowing downstream is x+y, thus the distance traveled is

d=(x+y).t_1

Where t1=2.5 hours. Substituting values:

20=(x+y)*2.5

Rearranging, we find the downstream equation:

2.5x+2.5y=20\qquad[1]

The speed when rowing upstream is x-y, and the distance traveled is

d=(x-y).t_2

Where t2=5 hours. Substituting values:

20=(x-y)*5

Rearranging, we find the upstream equation:

5x-5y=20\qquad[2]

Multiplying [1] by 2:

5x+5y=40

Adding this equation to [2]:

10x=60

Solving:

x=60/10=6

Dividing [2] by 5:

x-y=4

Solving for y

y=x-4=6-4=2

Thus, Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour

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egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

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2 × 2043231 = n² + 3·n + 2

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\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

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Learn more about the sum of a series here:

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Ax + By = C form of the given equation is –6x + y = –28.

Solution:

Given equation is y – 2 = 6(x – 5).

To write the equation in Ax + By = C format:

y – 2 = 6(x – 5)

y – 2 = 6x – 30

Add 2 on both sides of the equation.

y – 2 + 2 = 6x – 30 + 2

y = 6x – 28

Subtract 6x from both sides of the equation.

y – 6x = 6x – 28 – 6x

y – 6x = –28

Arrange the terms in the equation.

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Answer: k=\frac{3}{7}

Step-by-step explanation:

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