Answer:
The period of the graph of the traffic flow at 1st Ave and High St is half as long as the period of the graph of the traffic flow at 2nd Ave and Central.
Step-by-step explanation:
One way is subsitute
I like solving
so
what you do is this
add the equaitons to gether to eliminate x
2y=x+5
<u>3y=-x-3 +</u>
5y=0x+2
5y=2
y=2/5
sub back
2y=x+5
2(2/5)=x+5
4/5=x+5
minus 5
4/5-25/5=x
-21/5=x
the point is (x,y) or
(-21/5,2/5)
aprox
(-4.2,0.4)
I think they wanted you to read the garph you should have attached and guess the answer
the closes one is
hmm, x=-3
and y=1
so then D is the answer
Answer:
The limit of the function does not exists.
Step-by-step explanation:
From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.
For x<2

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.
For x>2

The value of y is 1 at x=2, because of he close circles on (2,1).
For x=2

Therefore the graph represents a piecewise function, which is defined as

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.



Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.
Answer:
you are correct :D
Step-by-step explanation: