The answer to this question is y = –3x^2, just had this on e2020 :)
There are 6/10 left over. 6/10 reduces to 3/5.
Step-by-step explanation:
2m+m=10+8
3m=18 /3
M=6
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:
63/5 or 12 (3/5)
Step-by-step explanation:
It is 8 whole (1 ÷ 5) plus 4 whole (2 ÷ 5)
So,we can write it as 8 (1/5) + 4 (2/5)
Then we will take those mixed fractions to improper fractions by using the rule of multiplication.
8 will be multiplied by 5 and then the value will be added to 1. Same rule will be applied for the second fraction.
(41÷5) + (22÷5)
Now we will take L.C.M. (Least common multiple) for both denominators 5 and 5. The L.C.M of 5 and 5 = 5
(41 + 22)/5
= 63/5
It is the answer. Also we can make this improper fraction into mixed fraction by dividing the whole number by 5.
12 (3/5)
