the answer is 3
but a way to work these problems out is to use PEMDAS
p-parentheses
e-exponent
m-multiplication
d-division
a-addition
s-subtraction
they have to be donr in that order or the answer will come out different
Answer:
6 Meters long
Step-by-step explanation:
3+3=6
12-6=6
Answer:
13.4
Step-by-step explanation:
Divide the number of meters in a mile (1,609.344) by the number of seconds in an hour (3,600): 1,609.344 divided by 3,600 equals 0.44704. One mile per hour equals 0.44704 meters per second.
If

is the amount of salt in the tank at time

, then the rate at which this amount changes over time is given by the ODE


We're told that the tank initially starts with no salt in the water, so

.
Multiply both sides by an integrating factor,

:




Since

, we have

so that the amount of salt in the tank over time is given by

After 10 minutes, the amount of salt in the tank is
Answer:
Yes you are correct.
Step-by-step explanation: