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3 years ago

10

The radius of the earth is 6378.1 km what is circumfrence of the earth in metres in standard form how many times can light encir

cle the earth per second

1 answer:

AVprozaik [17]3 years ago

8 0

Answer:

Two things we need to remember:

Speed of light = c = 299,792,458 m / s

This means that the light in one second covers a distance of 299,792,458 m

And for a circle of radius R, the is:

C = 2*pi*R

where:

pi = 3.14

Then if the radius of the earth is 6,378.1 km, then the of the earth is:

C = 2*(3.14)*6,378.1 km = 40,054.468 km

And we know that 1 km = 1000m

Then the circumference of the earth is:

C = (40,054.468*1000) m = 40,054,468 m

Now we need to see how many times the light can encircle the earth per second. (we assume that the light can move in a circular path, this is not really accurate to the reality)

We know that in one second the light can cover a distance of: 299,792,458 m

And the circumference of the earth is: 40,054,468 m

The quotient between these two distances is the number of times that the light can encircle the earth:

N = (299,792,458 m)/(40,054,468 m) = 7.5

So light could encircle the earth 7.5 times per second.

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What is the first step in solving: 14 + 2x - 3 = 43?

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Nora's leaf collection consisted of 150 leaves by August. By december , her collection had grown to 177 leaves. How much did her

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2 years ago

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)

S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

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3 years ago

Show work, please someone help me, i'm confused still on this concept.

BigorU [14]

Answer: choice C) -15x^4y

-----------------------------------------

Explanation:

The coefficients are -3 and 5. They are the numbers to the left of the variable terms
Multiply the coefficients to get -3*5 = -15. So -15 is the coefficient in the answer

Multiply the x terms to get x^3 times x = x^(3+1) = x^4. Notice the exponents are being added

Do the same for the y terms as well: y^2 times y^(-1) = y^(2+(-1)) = y^(2-1) = y^1 = y

So we have a final coefficient of -15, the x terms simplify to x^4 and the y terms simplify to just y

Put this all together and we end up with -15x^4y which is what choice C is showing

3 0

3 years ago