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3 years ago

Evan went to the park and saw four animals.

Mathematics

1 answer:

dsp733 years ago

4 0

Answer:

He saw 1 duck and 3 dogs.

Step-by-step explanation:

Given:

Evan went to the park and saw four animals.

Each animal was either a duck or a dog.

He saw a total of 14 legs.

Now, to find the number of each animal he had seen.

Let the number of duck be $x.$

And the number of dog be $y.$

So, total number of animals:

$x+y=4\\y=4-x$

Now, the total number of legs he did see:

As we know the legs of a duck are 2 and a dog are 4.

$2x+4y=14\\2x+4(4-x)=14\\2x+16-4x=14\\-2x+16=14$

Adding both sides by -16 we get:

$-2x=-2$

Dividing both sides by -2 we get:

$x=1$

The number of duck = 1.

Now, putting the value of $x$ in above equation we get:

$y=4-x$

$y=4-1\\y=3.$

The number of dogs = 3.

Therefore, he saw 1 duck and 3 dogs.

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4 years ago

Find the work required to move an object in the force field F = ex+y <1,1,z> along the straight line from A(0,0,0) to B(-1

storchak [24]

Answer:

Work = e+24

F is not conservative.

Step-by-step explanation:

To find the work required to move an object in the force field

$\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})$

along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.

Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with

r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1

r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t) with 0≤ t ≤ 1

is a parameterization of the segment.

the work W required to move an object in the force field F along the straight line from A to B is the line integral

$\large W=\int_{C}Fdr$

where C is the segment that goes from A to B.

$\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt$

Integrating by parts the last integral:

$\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1$

and

$\large \boxed{W=\int_{C}Fdr=e+24}$

To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24

Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.

The segment that goes from A to (1,0,0) can be parameterized as

r(t) = (t,0,0) with 0≤ t ≤ 1

so the work required to move the particle from A to (1,0,0) is

$\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1$

The segment that goes from (1,0,0) to B can be parameterized as

r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1

so the work required to move the particle from (1,0,0) to B is

$\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}$

Hence, the work required to move the particle from A to B along D is

e - 1 + (25e)/2 = (27e)/2 -1

since this result differs from e+24, the force field F is not conservative.

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Step-by-step explanation:

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