Incorrect Question 12 Calculate these complex numbers and express your results in rectangular form: 32/-20° 20 + (6-j8)(4+j2) -10+j24 -0.4151-j0.6281 -0.6281 0.4151j -0.6281 +0.4151j 0.4151-j0.6281

From step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?

thank you in advance

perform addition  -18+j23  and 30-j15              subtraction (8-j4)-3<25             multiplication  67+j84 and 102<40             division :  (28<30)/(14-j12)

Draw a seven-segment display, show the number highlighted in figure 1 (d3 -d0 are the input, a -g are the outputs). III. Simulation Wavetorm Editor - Cers/Ahmad Fainuz/Desanop/DLD Lab Practice/bedseg - beaeg - bcaseg 2018o03121936sim.vwwt Ee ep Ea w smutation search altera.com Haster Time Bar: ops Pointer: s29 us Start End iterval 529 us o ps 0 ps २ 2.56 us S12 us 75 us 10.24 us 12 s 15.36 us 172 us value at Name Ops do Figure 1

find the instantane ous expressions or the applied voltage V, voltage voltage figure. Als draw across resistos Vp and across aeross capaciter the phaser diengram. Vc in Ve R= 62 iz1o 5z sin (6250t)

The answers are in rectangular form

Calibri (Body) -11 by AaBbCcDx AaBbCcD AaBbC AaBbCc AaB AaBbCc. mat Painter BIU 1Normal 1No Spaci. Heading 1 Heading 2 Title Subtitle Chang Styles d Font Paragraph Styles IMPEDANCE QUIZ # 2 1. What capacitance when connected in series with a 500Q resistor will limit the current drawn from a 48-mV 465-kHz source to 20µA? * a. 144pF b. 145pF c. 146pF d. 147pF MY ANSWER: c.146pF 2. What is the total impedance at 20kHz of a series circuit consisting of a 1.5mH inductance, a 100Q resistance, and a 0.08uF capacitance? * a. 1340 41.7 b. 1930 L-37.2" C. 920 290 d. 530 2-12.6°

Find the 500K Rsh voltagea look Rang Rsh IG IAVO IG IAVE 0.03 7 70 525 500 600 350 3 7 70 700 262 To Imx Rm Rsh: I rang-Im 300 x31550 500-300 Ash 525 300x350 bo0 -300 Rsh = 350e %3D = 300X350 = 262n F00 - 300

What is (j205-j307+j47)-1? a. 1 b. -1 c. j d. -j