<span>To solve these GCF and LCM problems, factor the numbers you're working with into primes:
3780 = 2*2*3*3*3*5*7
180 = 2*2*3*3*5
</span><span>We know that the LCM of the two numbers, call them A and B, = 3780 and that A = 180. The greatest common factor of 180 and B = 18 so B has factors 2*3*3 in common with 180 but no other factors in common with 180. So, B has no more 2's and no 5's
</span><span>Now, LCM(180,B) = 3780. So, A or B must have each of the factors of 3780. B needs another factor of 3 and a factor of 7 since LCM(A,B) needs for either A or B to have a factor of 2*2, which A has, and a factor of 3*3*3, which B will have with another factor of 3, and a factor of 7, which B will have.
So, B = 2*3*3*3*7 = 378.</span>
The answer
by using fundamental definition
<span>The conditional probability of event B given event A is P(B|A)=P(A and B)/P(A) when two events are not independent.
so the only true answer is A, because </span>P(B|A)=P(A) if A and B are independents<span>
(definition)</span>
Answer:
hypotenuse length is 26
Step-by-step explanation:
a²+b²=c²
10²+24²=c²
100+576=c²
√676=26
Answer: b) f(m)
Step-by-step explanation:
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
