Answer:61.7
Step-by-step explanation:
maths
Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
-----------------
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.
================================================
Problem 8)
The answer is choice B) 5
-----------------
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
To factor (q+3)(q+6)
q^2+6q+3q+18
q^2+9q+18
(q+9)(q+2)
Answer:
6/10
Step-by-step explanation:
Answer:
The answer to your question is Advance = $35 and same-day = $40
Step-by-step explanation:
Advanced = a
Same-day = s
Equations
a + s = 75 ----- (I)
20a + 35s = 2100 ----- (II)
Process
1.- Solve the equations by elimination.
Multiply equation I by -20
-20a - 20s = -1500
20a + 35s = 2100
Simplify
0 + 15s = 600
Solve for s s = 600/15
s = $ 40
Substitute "s" in equation I to find "a"
a + 40 = 75
Solve for a
a = 75 - 40
a = 35
