Question

If you stand on a ship in a calm sea, then your height x (in ft) above sea level is related to the farthest distance y (in mi) that you can see by the equation:

y=sqrt(1.8x+(x/5280)^2)

(b) How high up do you have to be to be able to see 5 mi? (Round your answer to the nearest whole number.)

Answer 1

our height x (in ft) above sea level is related to the farthest distance y (in mi) that you can see by the equation:
\[y=\sqrt{1.8x+{(\frac{x}{5280})}^2}\]
here we have to find x for which we are able to see 5 miles, so put y=5
\[\large 5=\sqrt{1.8x+{(\frac{x}{5280})}^2}\]
square both sides and solve the quadratic for x, can you do it @Brent0423

Answer 2

ive already tried that and i get

1.8x+x^2+696960000

Answer 3

use the quadratic formula to solve for x
\[ax^2+bx+c=0\]
\[\large x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

Answer 4

-1.8+-sqrt((-1.8)^2-4(1)(696960000))/2(1)

Answer 5

im getting an imaginary number, this hw is due in 16 mins can u please just help me

Answer 6

let me check

Answer 7

\[25=1.8x+(\frac{x}{5280})^2\]
\[\frac{x^2}{5280^2}+1.8x-25=0\]
Use quadratic formula
\[\Large x=\frac{-1.8\pm\sqrt{(1.8)^2-4\times \frac{1}{5280^2}\times (-25)}}{2\times \frac{1}{5280^2}}\]
now it's just simplification
\[\Large x=\frac{-1.8\pm\sqrt{(1.8)^2+4\times \frac{1}{5280^2}\times 25}}{2\times \frac{1}{5280^2}}\]'
could you solve this from here?

Answer 8

i dont have calculator please tell me the answer

Answer 9

hurry only 3 mins remaining!

Answer 10

help!!!!!!

Answer 11

we'll get
\[\Large x=\frac{-1.8\pm \sqrt{3.24000358701}}{7.17401286\times 10^{-8}}\]

\[\Large x=\frac{-1.8\pm 1.80000099639}{7.17401286\times 10^{-8}}\]
we'll get two x
\[x=13.889, -5.00\times 10^{17}\]
the positive x is your answer

Answer 12

Did you understand @Brent0423 ?