Scientific notation, expodents, = one word problem

[Guest]

How do you do work the expodents here? I don't get it.

Halley's Comet orbits the sun about every 76 years. The comet travels in an elliptical path, with the sun at one of the foci. At the closest point, or perihelion, the distance of the comet to the sun is 8.8 * 10^7 km. At the furthest point, or aphelion, the distance of the comet from the sun is 5.3 * 10 ^9 km. Write an equation of the sllipse that models the path of Halley's Comet. Assume the sun is ont he x-axis.

So what I did was (8.8* 10^7 + 5.3 * 10^9 ) / 2 =a
but how do you do that with the expodents different???

-Anna

 

[soroban]

Hello, anna!

Halley's Comet orbits the sun about every 76 years.
The comet travels in an elliptical path, with the sun at one of the foci.
At the closest point, or perihelion, the distance of the comet to the sun is 8.8 * 10^7 km.
At the furthest point, or aphelion, the distance of the comet from the sun is 5.3 * 10 ^9 km.
Write an equation of the sllipse that models the path of Halley's Comet.
Assume the sun is ont he x-axis.

So what I did was (8.8* 10^7 + 5.3 * 10^9 ) / 2 =a
but how do you do that with the expodents different???

I'm surprised that you didn't reach for your calculator . . . almost everyone does!

The rest of them would simply write it out, like any addition problem:
. . . 88,000,000 + 5,300,000,000 .= . 5,388,,000,000
. . . Divide by 2: . 2,694,0000,000 . . . or: 2.694 x 10⁹


But since you ask ... there is a way to manipulate it:

. . . (8.8 x 10⁷) + (5.3 x 10⁹)

. . . (8.8 x 10⁷) + (5.3 x 10² x 10⁷)

. . . (8.8 x 10⁷) + (530 x 10⁷)

Factor: . (8.8 + 530) x 10⁷

Finally: . (538.8) x 10⁷ . = . 5.38 x 10⁹

But it's hardly worth the trouble, is it?