Answer:
3:4
Step-by-step explanation:
divide both sides by 6 to get simplest form
Answer:
ok so the two other ways are- 6:2 and 9:3
And the second part- 5
Step-by-step explanation:
Answer:
x = pi, 3pi, 5pi etc
x = -pi, -3pi, -5pi etc
Step-by-step explanation:
cos^2 x + 2 cos x + 1 = 0
replace cos x with m
m^2 + 2m +1 = 0
solve by factoring (this is a^2 +2ab+b^2 = (a+b)^2 where a =m and b=1)
(m+1)^2=0
take the square root of each side
m+1 =0
m=-1
now replace m with cos x
cos x = -1
take arccos of each side
arccos cos x = arccos (-1)
x = pi, 3pi, 5pi etc
Answer:
.
Step-by-step explanation:
OKAY, IM DOING A TEST WITH THIS QUESTION IM PRETTY SURE U NEED TO 4/23+5/14 AND WHATEVER THAT = YOU THEN SUBTRACT 8/12 AND I FORGOT HOW TO DO THE DARN ADDING-
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.
