Answer:
Rate of change = 3/2
Step-by-step explanation:
Use a function with an interval:
f(x) = x^2, [2,3]
<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>
Step-by-step explanation:
So first of all we plug in 1 into f(x) and the result of that into g(x).
f(1)=(1)^2-3(1)+5
=1-3+5
=3
g(3)=(3)(22)-2(3)
=66-6
=60
The circle equation is in the format (x – h)² + (y – k)² = r², with the center being at the point (h, k) and the radius being "r".
QUESTION 11.
Equation x²+y²+10x-14y-7 =0 can be rewritten as: x²+10x+25 + y² -14y + 49 -7 - 25 - 49=0
It can be factories as (x + 5)² + (y – 7)² = 9²
Therefore the radius equals 9 and the center is (-5,7)
QUESTION 12.
From equation (x + 4)² + y² = 121
The radius equals √121 = 11 and the center is (-4,0)
QUESTION 13.
As there are missing information in the question, I can't assist. However, you can use the general circle equation (x – h)² + (y – k)² = r² to solve the question.
Finally equations 14 & 15 aren't linear.
Hope that helps you :)
