Two pools are being drained. To start, the first pool had 3650 liters of water and the second pool had 4166 liters of water. Water is being drained from the first pool at a rate of 27 liters per minute. Water is being drained from the second pool at a rate of 39
liters per minute.
Let x be the number of minutes water has been drained.
(a) For each pool, write an expression for the amount of water in
the pool after x minutes.
O=D
х
Amount of water in the first pool (in liters) - ]
Amount of water in the second pool (in liters) = 1
5
?
(b) Write an equation to show when the two pools would have
the same amount of water.

b. x- 48 =120

step-by-step explanation:

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.