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skelet666 [1.2K]
3 years ago
14

A boat can travel 12 km down the river in the same time it can go 4 km up the river. If the current in the river is 2 km per hou

r, how fast can the boat travel in still water?
I know the answer is "4 km per hour" from the book but I don't know how to show the work so please help me. thank you
Mathematics
1 answer:
blondinia [14]3 years ago
4 0

Answer:

The speed of the boat in still water is 4 km per hour

Step-by-step explanation:

Let

s ---> the total speed of the boat in km/h

x ----> the speed of the boat in still water in km/h

t ----> the time in hours

d ---> the distance in km

Remember that the speed is equal to the distance divided by the time

so

The time is the distance divided by the speed

so

<em>Down the river</em>

t=\frac{d}{s}

we have

d=12\ km\\s=(x+2)\ km/h

Remember that the speed of the boat down the river is equal to the speed of the boat in still water plus the speed of the current

substitute

t=\frac{12}{x+2} ----> equation A

<em>Up the river</em>

t=\frac{d}{s}

we have

d=4\ km\\s=(x-2)\ km/h

Remember that the speed of the boat up the river is equal to the speed of the boat in still water minus the speed of the current

substitute

t=\frac{4}{x-2} ----> equation B

Equate equation A and equation B

\frac{12}{x+2}=\frac{4}{x-2}

solve for x

Multiply in cross

12(x-2)=4(x+2)\\12x-24=4x+8\\12x-4x=24+8\\8x=32\\x=4\ km/h

therefore

The speed of the boat in still water is 4 km per hour

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Evaluate.
lys-0071 [83]

Answer:

  361/900

Step-by-step explanation:

  \left(-\dfrac{1}{6}+0.6\left(-\dfrac{1}{3}\right)+1\right)^2=\left(-\dfrac{1}{6}-0.2+1\right)^2\\\\=\left(1-\left(\dfrac{1}{6}+\dfrac{1}{5}\right)\right)^2=\left(1-\dfrac{5+6}{6\cdot5}\right)^2=\left(\dfrac{19}{30}\right)^2=\boxed{\dfrac{361}{900}}

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2 years ago
Please help!! 20 POINTS
rjkz [21]

f(-3) would be 36.

When looking at synthetic division, the numbers across the top represent the coefficients of x^2, x and the constant in that order. Therefore, the equation is as follows.

2x^2 - 5x + 3

Now we can put -3 into the equation and solve.

2(-3)^2 - 5(-3) + 3

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7 0
3 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

3 0
3 years ago
Explain the meaning of (0,0) on a coordinate plane
Volgvan
(0,0) means the origin of the coordinate plane and its when bothe the x and y xoordinates are 0
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5) Jeff has a set monthly housing budget of $812. If Jeff is in an 18% tax bracket and his
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Answer:

2.406%

Step-by-step explanation:

28600*18%=5148

$5148-28600=$23,452

$23452/12 months=$1954.33 a month take home

1954.33/812=2.406%

6 0
3 years ago
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