Can anyone help me with this math problem?
1 answer:
In 1993, Moose Population: 3280
In 1999, population became: 4960
P (Population) , t (years)
t = 6 —> 4960 - 3280 = 1680
Average Change —> 1680/6 = 280 moose/year
• In terms of 1990:
t = 3 —> 3280-3 (280)
P(1990) = 2440
P(t) = 2440 + 280t
• In 2003; t = 13
P(13) = 2440 + 280 (13)
P(13) = 2440 + 3640
P(13) = 6080
• Moose population in 2003
= 6080
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Step-by-step explanation:
We are asked to write a system of equations using our given information.
Let x be number of small dogs and y be number of large dogs.
We have been given that paws at play made a total of $1,234 grooming 22 dogs, which means that number of small dogs and large dogs is 22. We can represent this information as:
We have been given that paws at play charges $43 to groom each small dog and $75 for each large dog. So the total grooming charges for grooming x small and y large dogs will be 43x+75y, that is equal to total charges for grooming $1234.
We can represent this information as:
Therefore, our desired system of equations is: