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To check for continuity at the edges of each piece, you need to consider the limit as approaches the edges. For example,
has two pieces, and
, both of which are continuous by themselves on the provided intervals. In order for
to be continuous everywhere, we need to have
By definition of , we have
, and the limits are
The limits match, so is continuous.
For the others: Each of the individual pieces of are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.
Answer:
<h3><em>D. 880 = 45d + 70; 18 days.</em></h3>Step-by-step explanation:
We are given fixed monthly charge = $70.
The cost of preschool per day = $45.
Number of days = d.
Total cost of d days = cost per day × number of days + fixed monthly charge.
Therefore, we get equation
880 = 45×d+70
<h3>880 = 45d +70.</h3>Now, we need to solve the equation for d.
Subtracting 70 from both sides, we get
880-70 = 45d +70-70
810=45d
Dividing both sides by 45, we get
18=d.
Therefore,<em> 18 days Barry attended preschool last month.</em>
<em>Therefore, correct option is D option.</em>
<h3><em>D. 880 = 45d + 70; 18 days.</em></h3>