Answer:
2
Step-by-step explanation:
Answer:
The correct option is;
B
Step-by-step explanation:
The given system of inequalities are;
5·x - 4·y > 4...(1)
x + y < 2...(2)
Representing both inequalities as a function of "y", gives;
For, 5·x - 4·y > 4...(1), we have;
-4·y > 4 - 5·x
y < 4/(-4) - 5·x/(-4)
∴ y < 5·x/4 - 1
For x + y < 2...(2), we have;
y < 2 - x
Therefore, y is less than the values given by the equation of the straight line equalities, and the feasible region is given by the common region under both dashed lines representing both inequalities as shown in the attached diagram created using Microsoft Excel
The correct option is therefore, B.
Factors and Multiples. But they both involve multiplication: Factors are what we can multiply to get the number. Multiples are what we get after multiplying the number by an integer (not a fraction).
Answer:
(a) true
(b) true
(c) false; {y = x, t < 1; y = 2x, t ≥ 1}
(d) false; y = 200x for .005 < |x| < 1
Step-by-step explanation:
(a) "s(t) is periodic with period T" means s(t) = s(t+nT) for any integer n. Since values of n may be of the form n = 2m for any integer m, then this also means ...
s(t) = s(t +2mt) = s(t +m(2T)) . . . for any integer m
This equation matches the form of a function periodic with period 2T.
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(b) A system being linear means the output for the sum of two inputs is the sum of the outputs from the separate inputs:
s(a) +s(b) = s(a+b) . . . . definition of linear function
Then if a=b, you have
2s(a) = s(2a)
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(c) The output from a time-shifted input will only be the time-shifted output of the unshifted input if the system is time-invariant. The problem conditions here don't require that. A system can be "linear continuous time" and still be time-varying.
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(d) A restriction on an input magnitude does not mean the same restriction applies to the output magnitude. The system may have gain, for example.
