If the price was $1 and it increased by 15% it would now be $1.15 therefore the multiplier is 1.15 as that increases a number by 15%
Answer:
a. 
b. 
c. 
Step-by-step explanation:
a. The volume of water initially in the fish tank = 15 liters
The volume of brine added per minute = 5 liters per minute
The rate at which the mixture is drained = 5 liters per minute
The amount of salt in the fish tank after t minutes = x
Where the volume of water with x grams of salt = 15 liters
dx = (5·c - 5·c/3)×dt = 20/3·c = 

b. The amount of salt, x after t minutes is given by the relation




c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;




Oh ok so the answer is .666666666666 etc infinite 6s
Answer:
r = -1 or r = 13/3
Step-by-step explanation:
Solve for r over the real numbers:
abs(10 - 6 r) = 16
Split the equation into two possible cases:
10 - 6 r = 16 or 10 - 6 r = -16
Subtract 10 from both sides:
-6 r = 6 or 10 - 6 r = -16
Divide both sides by -6:
r = -1 or 10 - 6 r = -16
Subtract 10 from both sides:
r = -1 or -6 r = -26
Divide both sides by -6:
Answer: r = -1 or r = 13/3
Answer:
Step-by-step explanation:
and
and we are told that C is 6cm longer than A. That means that C = A + 6.
We are going to cross multiply each one of those ratios. The first one gives us
4A = 3B and the second one gives us
9B = 8C. But since C = A + 6, then
9B = 8(A + 6) and
9B = 8A + 48 and
Now we will solve the first equation above for A:
If 4A = 3B, then
and will use that as a sub for A in the second equation:
and
9B = 6B + 48 and
3B = 48 so
B = 16.
Now that we know B, we can use it to solve for A:
4A = 3(16) and
4A = 48 so
A = 12.
Then we can use that all the way back in the expression we set up for C:
C = A + 6 so
C = 12 + 6 so
C = 18
12 + 16 + 18 is the length of the string: 46cm