Find the x and y-intercepts of the equation. Then calculate the points at constant intervals and plot and connect those points on a makeshift graph. A quadratic equation can have 2 solutions, 1 solution, or none. The corresponding graphs(if two solutions) would most probably cross over the x-axis twice.
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Answer:
In 1981 was the building worth double it’s value.
Step-by-step explanation:
Given : A townhouse in San Francisco was purchased for $80,000 in 1975. The appreciation of the building is modeled by the equation :
, where t represents time in years.
To find : In what year was the building worth double it’s value in 1975?
Solution :
The amount is $80,000.
The building worth double it’s value in 1975.
i.e. amount became A=2(80000).
Substitute in the model,



Taking log both side,



i.e. Approx in 6 years.
So, 1975+6=1981
Therefore, in 1981 was the building worth double it’s value.
Answer:
6.25 hours
Step-by-step explanation:
As we know that,
1 hour = 60 minutes
⇒ 
⇒ 
⇒ 375 mintes = 6.25 hours
Thus, 375 mintes = 6.25 hours