A water well drilling rig has dug to a height of -60 feet after one full day of continuous use. 1. Assuming the rig drilled at a constant rate, what was the height of the drill after 15 hours?

2. If the rig has been running constantly and is currently at a height of -147.5 feet, for how long has the rig been running?

1.677

step-by-step explanation:

i'm not good at explaining stuff but if you need an answer i'm here for you.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

hope this ^_^

3 qrs + 3/2 xy hope this : /

step-by-step explanation: