Techically bananas are nuts...
The numerator factors as (t-8)(t+4), so the whole thing is (t-8)(t+4)/(t-8). Now we can cancel the t-8 AS LONG as t isn't equal to 8, otherwise, we are canceling a zero from both numerator and denominator which is invalid. What is left is t+4.
So your choice was right.
Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10
Answer:
Domain is that interval in which the function is defined ....
So this function is defined everywhere except 4 and 7 ,
option D is correct....
Answer:
78 units squared
Step-by-step explanation:
