The solution is sometimes integers
One.
Two sides are the same, the base is 1m.
Answer:
- <u>No, he can get an output of 0 with the second machine (function B) but he cannot get an output of 0 with the first machine (function A).</u>
Explanation
The way each machine works is given by the expression (function) inside it.
<u>1) </u><em><u>Function A</u></em>
To get an output of 0 with the function y = x² + 3, you must solve the equation x² + 3 = 0.
Since x² is zero or positive for any real number, x² + 3 will never be less than 3 (the minimum value of x² + 3 is 3). So, it is not possible to get an output of 0 with the first machine.
<u>2) </u><em><u>Function B</u></em>
Solve 
So, he can get an output of 0 by using x = 4.
Answer:
Emily receives $1,750 less after the shift
Step-by-step explanation:
9514 1404 393
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.