Answer:
turning it I think.........
Answer:
The sale price of the item is $
255
Step-by-step explanation:
To determine the amount of discount (
d
) offered by the store, we calculate
15
% of the sale price. Thus:
d = 300 × 15
÷ 100
d = 3
00 × 15 ÷ 1
00 (take the 2 zeros out on both the hundred and 3 hundred)
d = 3 × 15
d = 45
If the amount of discount is known, the sale price (
s
) of the item will be the regular price minus the discount. Hence:
s = 300 − 45
s = 255
Hope this helps you! If not sorry!
Answer:Solve this system by substitution (3 variables) brainliest answer goes to first answer and 8 points!
-x -y -z = -8
-4x +4y +5z = 7
2x+2z=4
Show steps please!
Step-by-step explanation: a-z=3-2=657u568959jb5j656856 thats the answer your welcome
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:
8 new chicks can be fitted in the coop.
Step-by-step explanation:
A chicken coop holds 10 hens or 20 chicks.
That means space in the coop for 10 hens = space for 20 chicks
Or space for 1 hen = space for 2 chicks
Now
th of the hens were removed from the coop.
So, number of hens removed from the coop = 
= 4 hens
And space for 4 hens in the coop = space for the 8 chicks
Therefore, 8 new chicks can be fitted in the coop.