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3 years ago

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a boat makes a 120-mile trip downstream in 3 hours but makes the return trip in 4 hours. if b=the rate of the boat in still wate

r and c=the rate of the current, which of the following equations represents the trip upstream?

1 answer:

ASHA 777 [7]3 years ago

8 0

it would be equation c, since you are you going upstream and the current would be going down stream

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018 10.0 points

Elina [12.6K]

Answer:

what grade

Step-by-step explanation:

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2 years ago

(1,4,9,4,10,8,42)

mixas84 [53]

The answer is has to be D

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2 years ago

Find x<br> A. 22√2<br> B. 22<br> C. 11√3/3<br> D. 11√3

mel-nik [20]

Answer

$22\sqrt{2}$

Step-by-step explanation:

First we gave a name to the side that is common for both triangles, can be Y for example.

Using trigonometric functions we can make a relationship between the two angles that we have and the magnitud of the side.

$sin(45)=\frac{11}{Y} \\$ , Because sin(x)= op/hip

$cos(60)=\frac{Y}{x}$ , Because con(x)=ad/hip

Having this equations, we can solve the system and find x

$Y=\frac{11}{sin(45)} \\Y=cos(60)*x$

we equal Y and solve

$\frac{11}{sin(45)}=cos(60)*x\\ x=\frac{11}{sin(45)*cos(60)} \\x=\frac{11}{\frac{\sqrt{2} }{2} *\frac{1}{2} } \\x=\frac{11}{\frac{\sqrt{2} }{4} } \\x=\frac{44}{\sqrt{2} }\\$

Then we rationalize

$x=22\sqrt{2}$

Hope you like it!

7 0

2 years ago

Read 2 more answers

djyliett [7]

Answer:

Step-by-step explanation:

A = (3.14)(8)2

= 200.96

3 0

2 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago