Well, a yard is equal to 36 inches.
One foot and three inches is equal to 15 inches.
36 - 15 = 21
Hope this helps! :D
Answer: The answer is NO.
Step-by-step explanation: The given statement is -
If the graph of two equations are coincident lines, then that system of equations will have no solution.
We are to check whether the above statement is correct or not.
Any two equations having graphs as coincident lines are of the form -
If we take d = 1, then both the equations will be same.
Now, subtracting the second equation from first, we have
Again, we will get the first equation, which is linear in two unknown variables. So, the system will have infinite number of solutions, which consists of the points lying on the line.
For example, see the attached figure, the graphs of following two equations is drawn and they are coincident. Also, the result is again the same straight line which has infinite number of points on it. These points makes the solution for the following system.
Thus, the given statement is not correct.
Answer:
Step-by-step explanation:
Using the mean concept, it is found that the average score of Adrian’s favorite team relative to its opponent using the observations Adrian made during the game is of -0.3333.
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The mean of a data-set is given by the sum of all observations divided by the number of observations.
Adrian looked the scores every 10 minutes in 2 hours, thus 12 times, as .
4 times ahead by 3 points, thus
4 times behind by 4 points, thus
At the other moments, the game was tied, thus the observation is 0.
Thus, the mean is of:
The average score of Adrian’s favorite team relative to its opponent using the observations Adrian made during the game is of -0.3333.
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Answer:
- ABHGEFDCA
- does not exist
- ECBADFE
Step-by-step explanation:
A Hamiltonian circuit visits each node once and returns to its start. There is no simple way to determine if such a circuit exists.
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For graph 2, if there were a circuit, paths ACB, ADB, and AEB would all have to be on it. Inclusion of all of those requires visiting nodes A and B more than once, so the circuit cannot exist.
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For graph 3, the circuit must include paths BAD and DFE. That only leaves node C, which can be reached from both nodes B and E, so path ECB completes the circuit.
No....if you set them up as proportions and cross multiply the sides do not equal