Answer:
y=2x+1, assuming my change in the reported data was correct.
Step-by-step explanation:
The data for x had one more entry than the values for y. I removed the second "0" so that the x and y points line up, as shown in the attached image. The data indicate a straight line, with a slope of 2 (y increases by 2 for every x increase of 1). The y-intercept is 1, as per the first data point (0,1).
The answer is
X= -3/2 or x=1
Sin(B) = opp/hyp = 32/40 = 0.8
cos(B) = adj/hyp = 24/40 = 0.6
tan(B) = opp/adj = 32/24 = 1.33 (repeating)
The frequency at which the filter will be resonant is 1. 59 × 10^6 Hz
<h3>How to determine the frequency</h3>
The resonant frequency of the series RLC circuit is given as;
Frequency = 1/2π√(LC)
It means that the current will peak at the resonant frequency for both inductor and capacitor.
Substitute the values
Frequency = 1/2 × 3. 142 √ (50 × 10^-3 × 2 × 10^ -6)
Frequency = 1/2 × 3. 142 × 1 × 10 ^-7
Frequency = 1/ 6. 284 × 10^-7
Frequency = 1. 59 × 10^6 Hz
Thus, the frequency at which the filter will be resonant is 1. 59 × 10^6 Hz
Learn more about frequency here:
brainly.com/question/9324332
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