Question

tell me how to solve this.. plss

For altitudes h up to 10,000 meters, the density D of Earth's atmosphere (in kg/m^3) can be approximated by the formula

D = 1.225−(1.12 x 10^−4)h +(3.24 x 10^−9)h^2.

Approximate the altitude if the density of the atmosphere is 0.7 kg/m^3. (Round your answer to the nearest meter.)

Answer 1

plug in 0.7 for D, and then solve for h, you will have a quadratic equation

Answer 2

Pluging in we have
0.7 = 1.225−.000112*h +.00000000324*h^2.

Answer 3

im having misunderstandings about h and h^2..? can i just perform the operation or i need to do something..?

Answer 4

h is the unknown, and this is a quadratic equation. you will have to set it equal to zero and use quadratic formula

Answer 5

yep already did that.. after that i dont know what is the next step..?

Answer 6

ahhh.. wait.. h and h^2 cant be equated..? like h+h^2 = 2h^2..?

Answer 7

0.7 = 1.225−.000112*h +.00000000324*h^2.
subtract 0.7 from both sides
As a quadratic it should look like this :
.00000000324*h^2 −.000112*h +1.225 - 0.7 = 0

Answer 8

then after that..? what now..? it became 3.24x10^9h^2-.000112h+0.525.. whats next..?

Answer 9

use the quadratic formula

Answer 10

oh.. i get the point now.. haha.. sorry.. hwahah.. forgot to use that quadratic formula.. i've focused on the wrong thing solving for H by just deriving..

Answer 11

http://www.wolframalpha.com/input/?i=0.7+%3D+1.225%E2%88%92%281.12+x+10^%E2%88%924%29*h+%2B%283.24+x+10^%E2%88%929%29*h^2

Answer 12

wow.. is that another website just like mathway..? haha..

Answer 13

:)

Answer 14

remember the domain of this problem , 10,000 meters maximum

Answer 15

so we can ignore the 28,000 meter solution

Answer 16

the 28,975 solution

Answer 17

ahh.. so the first one is the answer because of the 10,000.. because 28,00 is way above it.. ahh.. tnx again.. xDDDD