Answer:
-7 is bigger
-9 is bigger
-6 is bigger
7 is bigger
I think.....
-9,-8,-7,-6,-4
-9,-7,-6,4,7
-2-1,0,4,9
-8,-2,3,5,8 ?
I guess I didn't understand the question
The problem requires us to find the rate or speed of Lilli in miles per hour. Therefore, we just have to divide the given data.
12/110 = 0.10909.... Therefore, Lilli's speed or rate is 0.11 mi/hr.
Answer:
c = -1
Step-by-step explanation:
3(c+5)=12
3 × c = 3c
3 × 5 = 15
3c + 15 = 12
<u> - 15 -15</u>
3c = -3
c = -3 ÷ 3 = -1
Check:
3(-1+5)=12
-3 + 15 = 12
Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min
Answer:
Step-by-step explanation:
Decrease = 100 - 49 = 51
Percentage of decrease =

= 51%