Mark is draining his pool to clean it. at 12: 00 p. m., he started to drain the pool. at 4: 00 p. m., the pool had 11,600 gallons of water. at 7: 00 p. m., the pool had 6,800 gallons of water. if draining the pool is modeled by a linear function, how many gallons of water were initially in the pool? a) 15,600 b) 18,000 c) 19,400 d) 21,200

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

y=mx+b

step-by-step explanation:

m is slope, b is y-intercept

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation: