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Arisa [49]
3 years ago
12

A small island is 5 miles from the nearest point p on the straight shoreline of a large lake. if a woman on the island can row a

boat 2 miles per hour and can walk 3 miles per hour, where should the boat be landed in order to arrive at a town 7 miles down the shore from p in the least time
Mathematics
1 answer:
Luba_88 [7]3 years ago
8 0
Time =5mile/hour/5mile=1 hour
then for 7 miles, the boat should be landed in the 4miles away from point p because of 7 -3 = 4 miles of being on a boat.
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Determine the initial height of the ball and the time interval before the ball hits the ground. initial height = 0; hits the gro
Anestetic [448]

Answer:

Answer: B) initial height = 150; hits the ground between 5 and 6 seconds

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Solve:<br> -8r - (-2r-3) = 13<br><br> (if you can give a step by step, that would be great!)
coldgirl [10]

Answer:

r =  -  \frac{5}{3}  = 1 \frac{2}{3}  =  - 1.6

Step-by-step explanation:

- 8r -  ( - 2r - 3) = 13

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- 6r + 3 = 13

- 6r = 13 - 3

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\boxed{\green{r =  -  \frac{5}{3}  = 1 \frac{2}{3}  =  - 1.6}}

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3 years ago
Solve for all real values of x.<br> x^2+6=0
irina [24]

Answer:

-6+6=0

x^2=-6

sq root of -6 = x

Step-by-step explanation:

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2 years ago
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Subtract -13xy from -5xy so<br> -5xy -(-13xy)=?
S_A_V [24]

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8xy

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Integrate this two questions ​
tankabanditka [31]

Simplify the integrands by polynomial division.

\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)

\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)

Now computing the integrals is trivial.

5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

8.

\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}

using the same approach as above.

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2 years ago
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