Answer: 
Step-by-step explanation:
Let x be the average number of pounds Fido must loss.
Since, the initial weight of Fido is 35 pounds.
And, After losing the weight, the new weight of Fido in pounds = 28 pounds.
Then the time taken for losing the weight
= 
= 
According to the question, it must lose weight within 6 months,
Thus, 
Which is the required inequality to find the average number of pounds per month.
By solving it we, get, 
The population of a town has approximately doubled every 17 years since 1950.
the equation P=
where Po is the population of the town in 1950, is used to model the population, P, of the town t years after 1950.
When t=17 yrs
P=2
for 1 year
The equation becomes
P =
-------------(1)
Our original equation is
----------------------------------(2)
equating expression 1 and 2

Cancelling
from both sides we get

t/17=k
⇒k=t/17 is the solution.
What is it supposed to go from
The answer is C and here are the steps
For 15 Hours Of Work, You Will Be Paid $157.50. Just Multiply 10.50 By 15 And You Got Your Answer.