Ask question

Ask question

All categories

3 years ago

6

A boat travels 33 miles downstream in 4 hours. The return trip takes the boat 7 hours. Find the speed of the boat in still water

.

2 answers:

Paha777 [63]3 years ago

6 0

<u>Answer:</u>

Speed of the boat in still water = 6.125 miles/hour

<u>Step-by-step explanation:</u>

We are given that a boat travels 33 miles downstream in 4 hours and the return trip takes the boat 7 hours.

We are to find the speed of the boat in the still water.

Assuming $S_b$ to be the speed of the boat in still water and $S_w$ to be the speed of the water.

The speeds of the boat add up when the boat and water travel in the same direction.

$Speed = \frac{distance}{time}$

$S_b+S_w=\frac{d}{t_1}=\frac{33 miles}{4 hours}$

And the speed of the water is subtracted from the speed of the boat when the boat is moving upstream.

$S_b-S_w=\frac{d}{t_2}=\frac{33 miles}{7 hours}$

Adding the two equations to get:

$S_b+S_w=\frac{d}{t_1}$

+ $S_b-S_w=\frac{d}{t_2}$

___________________________

$2S_b=\frac{d}{t_1} +\frac{d}{t_2}$

Solving this equation for $S_b$ and substituting the given values for $d,t_1, t_2$ :

$S_b=\frac{(t_1+t_2)d}{2t_1t_2}$

$S_b=\frac{(4 hour + 7hour)33 mi}{2(4hour)(7hour)}$

$S_b=\frac{(11 hour)(33mi)}{56hour^2}$

$S_b=6.125 mi/hr$

Therefore, the speed of the boat in still water is 6.125 miles/hour.

EleoNora [17]3 years ago

3 0

Answer:

$6.48\frac{mi}{h}$

Step-by-step explanation:

Let' call "b" the speed of the boat and "c" the speed of the river.

We know that:

$V=\frac{d}{t}$

Where "V" is the speed, "d" is the distance and "t" is the time.

Then:

$d=V*t$

We know that distance traveled downstream is 33 miles and the time is 4 hours. Then, we set up the folllowing equation:

$4(b+c)=33$

For the return trip:

$7(b-c)=33$ (Remember that in the return trip the speed of the river is opposite to the boat)

By solving thr system of equations, we get:

- Make both equations equal to each other and solve for "c".

$4(b+c)=7(b-c)\\\\4b+4c=7b-7c\\\\4c+7c=7b-4b\\\\11c=3b\\\\c=\frac{3b}{11}$

- Substitute "c" into any original equation and solve for "b":

$4b+\frac{3b}{11} =33\\\\4b+\frac{12b}{11}=33\\\\\frac{56b}{11}=33\\\\b=6.48\frac{mi}{h}$

You might be interested in

Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient.

Shtirlitz [24]

Answer:

odd degree and positive leading coefficient

Step-by-step explanation:

From the graph , we can see that when x goes to infinity , y goes to infinity

As x--> ∞, y--> ∞ (As x increases the value of y increases on the positive side)

we can see that when x goes to -infinity , y goes to -infinity

As x--> -∞, y--> -∞ (As x decreases the value of y decreases on the negative side)

When x--> ∞, y--> ∞ and x--> -∞, y--> -∞

The leading coefficient is positive and largest exponent is odd

So the graph has odd degree and positive leading coefficient

8 0

3 years ago

12-1/5r = 2r +1 solve for r

levacccp [35]

Answer:

r=5

Step-by-step explanation:

1.simplify both sides of the equation.

2.substract 2r from both sides.

3.substract 12 from both sides.

4.multiply both sides by 5/(11)

5. your answer will be 5.

r=5

6 0

3 years ago

Read 2 more answers

Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

Tonja is a married taxpayer who claims 2 withholding allowances. She earns

viva [34]

The amount that Tonja will save in Federal Income and FICA Taxes if she has her childcare expenses deducted from her check before taxes is: $26.17

<h3>What is Federal Income Tax?</h3>

Federal Income Tax is a tax that is levied by the Federal Government and most states in the United States of America. It comprises of 7 tax rates.

<h3>What is FICA Tax?</h3>

FICA Tax is a Federal Insurance Contributions Act Tax. This tax is levied on every payroll of employees to fund Social Security and Medicare programs. This is divided into two parts:

Social Security Tax which is 6.2% and
Medicare Tax which is 1.45%. The total is 7.65%

How to calculate Tonja's Tax before deduction

Federal Tax rate first level is 10% on 0$ - $19,900.

1) Hence, the first tax is 10% on $600, which is $60

2) The next tax is FICA. This is applied to the Gross Pay of $600.

Applying both taxed under FICA we have 7.65% * $600 = $45.9

Total Tax before a deduction is:

$60 + $45.9 = $105.6

The Tax Payable if childcare expenses is deducted first will be:

10% on $450 = $45

7.65% on $450 = $34.43

Total = $79.43

Hence the tax savings will be:

105.6 - 79.43 = $26.17

Learn more about FICA Tax at:
brainly.com/question/3214345

3 0

2 years ago

Select the two values of x that are roots of this equation x^2-5x+2=0

Nataly [62]

The roots of the equation is x = 4.56 OR x = 0.44

<h3>Quadratic equation</h3>

From the question, we are to determine the roots of the given equation

The given equation is

x² -5x +2 = 0

Using the formula method,

$x =\frac{-b \pm \sqrt{b^{2}-4ac } }{2a}$

In the given equation,

a = 1, b = -5, c = 2

Putting the values into the formula,

$x =\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(2) } }{2(1)}$

$x =\frac{5 \pm \sqrt{25-8} }{2}$

$x =\frac{5 \pm \sqrt{17} }{2}$

$x =\frac{5 + \sqrt{17} }{2}$ OR $x =\frac{5 - \sqrt{17} }{2}$

$x =\frac{5 + 4.12}{2}$ OR $x =\frac{5 - 4.12}{2}$

$x =\frac{9.12}{2}$ OR $x =\frac{0.88}{2}$

x = 4.56 OR x = 0.44

Hence, the roots of the equation is x = 4.56 OR x = 0.44

Learn more on Quadratic equation here: brainly.com/question/8649555

#SPJ1

3 0

2 years ago