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Anni [7]
3 years ago
9

A motorboat has a four hour supply of gasoline. How far from the marina can it travel if the rate going out against the current

is 20mi/h and the rate coming back with the current is 30 mi/h
Mathematics
1 answer:
Igoryamba3 years ago
5 0

Answer:

The distance the marina can travel = x_{} = 48 miles

Step-by-step explanation:

The rate going out against the current is = u - v = 20 \frac{mi}{h}

The rate going out with the current is = u + v = 30 \frac{mi}{h}

Let the distance traveled by the motorboat = x_{}

Total time the boat can travel = 4 hours

⇒ Total time = ( Time taken by boat to go against the current ) + ( Time                                  taken by boat to go  with the current )

⇒ 4 = \frac{x}{u - v} + \frac{x}{u + v }

⇒ 4 = \frac{x}{20} + \frac{x}{30}

⇒ 4 = \frac{x}{12}

⇒ x_{} = 48 miles

This is the distance the marina can travel.

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Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
40 times 40 times 30 equals
Fudgin [204]

48000 is your awnserrrr

5 0
3 years ago
The populations of two cities after t years can be modeled by -150t+50,000 and 50t+70,0000 . What is the difference in the popul
Sindrei [870]

Answer:

25,800

Step-by-step explanation:

Plug in 4 everytime you see T.

multiple-150 times 4 plus 50000 which gives you 49,400.

then multiple 50 times 4 and add 75000, giving you 75,200 then subtract 75,200 - 49,400 which you end up with 25,800.

hope this helped

5 0
2 years ago
On a piece of paper, use a protractor to construct a triangle with angle measures of 40° and 60º.
andrew-mc [135]

Answer:

There are 180 degrees in a triangle so it is 80 degrees

4 0
2 years ago
Read 2 more answers
How to solve this equation 5x-4-7x=4
elixir [45]

Answer:

x= -4

Step-by-step explanation:

Simplifying

5x + -4 = 7x + 4

Solving

-4 + 5x = 4 + 7x

Move all terms containing x to the left, all other terms to the right.

Add '-7x' to each side of the equation.

-4 + 5x + -7x = 4 + 7x + -7x

Combine like terms: 5x + -7x = -2x

-4 + -2x = 4 + 7x + -7x

Combine like terms: 7x + -7x = 0

-4 + -2x = 4 + 0

-4 + -2x = 4

Add '4' to each side of the equation.

-4 + 4 + -2x = 4 + 4

Combine like terms: -4 + 4 = 0

0 + -2x = 4 + 4

-2x = 4 + 4

Combine like terms: 4 + 4 = 8

-2x = 8

Divide each side by '-2'.

x = -4

Simplifying

x = -4

3 0
3 years ago
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