1.4 / 7
multiply by 10 in both up and down
(1.4*10) / (7*10)
14 / 70
divide by 7
(14/7) / (70/7)
2 / 10
divide by 2
(2/2) / (10/2)
1 / 5
<span>30 hours
For this problem, going to assume that the actual flow rate for both pipes is constant for the entire duration of either filling or emptying the pool. The pipe to fill the pool I'll consider to have a value of 1/12 while the drain that empties the pool will have a value of 1/20. With those values, the equation that expresses how many hour it will take to fill the pool while the drain is open becomes:
X(1/12 - 1/20) = 1
Now solve for X
X(5/60 - 3/60) = 1
X(2/60) = 1
X(1/30) = 1
X/30 = 1
X = 30
To check the answer, let's see how much water would have been added over 30 hours.
30/12 = 2.5
So 2 and a half pools worth of water would have been added. Now how much would be removed?
30/20 = 1.5
And 1 and half pools worth would have been removed. So the amount left in the pool is
2.5 - 1.5 = 1
And that's exactly the amount needed.</span>
If you draw only 25 balls, you could draw the 25 odd-numbered balls. However, there will then be no odd-numbered balls left, so when you draw two more you will be guaranteed to get two even-numbered balls. Thus, the minimum is 25+2=27 balls.
Answer:
the graph of the function is non-linear
the table can be represented by the equation y=4x+4
Step-by-step explanation:
non linear because 1+7=8 and then 10+2 = 12 so they have different numbers to add each time
represented by the equation because the x numbers (1,2,3,4,) all multiplied by 4 and added by 4 gives you y
any more help just ask :)
