Answer:
according to the question value got less 12% each year
first of all the 12% of 29000$ = 29000÷100×12= 3480 $
in one year truck value deprecates 3480$
in 10 years it will 3480$×10 =34800$
now truck cost will be = 29000-34800$= -5000$
Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
___
The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
4 to 7 = 4/7
5/8 is not.
12/20 = 6/10 = 3/5 . It is not equivalent.
8:14 = 4 : 7 . It is equivalent
5 to 10 = 1 to 2. It is not.
16/28 = 8/14 = 4/7. It is equivalent.
9 to 16 = 9 to 16. It is not equivalent.
So only 8:14 and 16/28 are equivalent to 4 to 7.
Answer:
Step-by-step explanation:
c+7
you have 7 more then a number c
