All categories

3 years ago

The distance from Earth to the sun is about 9.3 × 10 7th power miles.

Mathematics

1 answer:

AURORKA [14]3 years ago

5 0

Answer:

The distance from earth to sun is 387.5 times greater than distance from earth to moon.

Solution:

Given, the distance from Earth to the sun is about $9.3 \times 10^{7} \mathrm{miles}$

The distance from Earth to the Moon is about $2.4 \times 10^{5} \mathrm{miles}$

We have to find how many times greater is the distance from Earth to the Sun than Earth to the Moon?

For that, we just have to divide the distance between earth and sun with distance between earth to moon.

Let the factor by which distance is greater be d.

$\text { Now, } \mathrm{d}=\frac{\text { distance between sun and earth }}{\text { distance between moon and earth }}=\frac{9.3 \times 10^{7}}{2.4 \times 10^{5}}=\frac{9.3}{2.4} \times 10^{7-5}\\\\=3.875 \times 10^{2}=387.5$

Hence, the distance from earth to sun is 387.5 times greater than distance from earth to moon.

You might be interested in

Which ordered pair is a solution to this equation?

atroni [7]

I hope this helps you

4 0

3 years ago

Read 2 more answers

What is the quotient of 1/10 and 9/10

ira [324]

Answer:

1/9 = 0.1111111

Step-by-step explanation:

4 0

3 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago

The weight of Lyle’s dog increased from 29.7 pounds to 48.9 pounds in an 8-month period. What was the dog’s average weight incre

krok68 [10]

First subtract the weight the dog weighs now to the weight he was:

48.9
- 29.7
------------
19.2

Then take the number you got which is the weight the dog gain in total, and divide it by the months the dog gained weight to get the average weight gained each month:

19.2/8 = 2.4

The average weight gain each month was 2.4 pounds.

Hope this helped you!

6 0

3 years ago

4(x+3)-4=8(1/2x+1) i need to find the solution

oee [108]

You can solve this by algebraically as well as graphically. When I tried both ways I got:
solution is x= -2

5 0

3 years ago