![4G 7:27 AM 0.1KB/s Vo LTE LTE 82 III 14 1/2 21/3 T1/3 51/6 f/6. TT 1 7/B 1111/6 41/3 51/3 + 1 w p=sin(n:0) Х 0 <O< 27 2 Х n =](//img.homeworklib.com/questions/91235770-9311-11eb-824f-a38cb8868f68.png?x-oss-process=image/resize,w_560)
![4G 7:27 AM 0.OKB/s vo LTE LTE 82 III 1+ 1/2 21/3 T[/3 3 511/6 TT/ TT 1 711/6 1111/6 41/3 51/3 + 1 Х r = sin(n:0) <O< 7 0 2 Х](//img.homeworklib.com/questions/92294de0-9311-11eb-9269-5db97943e671.png?x-oss-process=image/resize,w_560)
![4G 7:29 AM|0.OKB/s Vo LTE LTE 82 III 21/3 T/3 511/6 TT/ TT 1 AT/6 1116 41/3 51/3 + 1 Х r = sin(n·0)| 0 <o< 27 2 x n = 3 12 -1](//img.homeworklib.com/questions/9408ad20-9311-11eb-98bc-4db510f48766.png?x-oss-process=image/resize,w_560)
![4G 7:29 AM|1.5KB/S Vo LTE LTE 82 III 14 1/2 21/3 T1/3 51/6 f/6. TT 1 7/6 1 1/6 41/3 517/3 + 1 Х r = sin(n·0)| 0 <o< 27 2 n =](//img.homeworklib.com/questions/949992e0-9311-11eb-bc0f-8b2e185e9d5f.png?x-oss-process=image/resize,w_560)
![4G 7:31 AM|1.7KB/s vo LTE LTE 82 4 2 1/2 21/3 13 51/6 TT/6 T 0 2 2 4 ZTt/6 111/6 41/3 51/3 + 1 N 7=1c sin(n :0) Х 0 <O< 26 2](//img.homeworklib.com/questions/9538d9e0-9311-11eb-8dd0-a317b758e2c8.png?x-oss-process=image/resize,w_560)
![4G 7:31 AM|0.0KB/s Vo LTE LTE 82 4 211/27 21/3 TT/3 5/6 TT/6 TT 0 2 2 4 ZTt/6 111/6 41/3 51/3 + 1 N 7=1c sin(n+ :0) Х 0 <O< 2](//img.homeworklib.com/questions/95c7ed40-9311-11eb-8553-efc39af5d396.png?x-oss-process=image/resize,w_560)
![4G 7:33 AM | 10.2KB/S Vo LTE LTE 81 4 211/27 21/3 T1/3 5ſt/6 TT/6 2 2 4 71/6 1111/6 41/3 51/3 + 1 Х p=1+c sin(n.6 :0) 0 <o< 2](//img.homeworklib.com/questions/965c1460-9311-11eb-9b4e-11d3b39e1960.png?x-oss-process=image/resize,w_560)
![4G 7:33 AM 0.0KB/s Vo LTE LTE 81 4 21/22 8/3 TT/3 51t/6 T/6 2 2 4 ZTt/6 111/6 41/3 Бт/3 + « 1 Х p=1+c sin(n.6 :0) 0 <o< 27 2](//img.homeworklib.com/questions/96f2be00-9311-11eb-bc30-e90f21598eca.png?x-oss-process=image/resize,w_560)
![4G 7:36 AM | 0.OKB/s Vo LTE LTE 81 III 101 T1/2 2/3 TT/3 51/6 TT/6 M 0 20 10 10 20 71/6 11T1/6 + 1 Х p= (1 - a cos(0)) os(0))](//img.homeworklib.com/questions/978be5b0-9311-11eb-818e-ad30ff7a57b9.png?x-oss-process=image/resize,w_560)
![4G 7:39 AM | 4.8KB/s Vo LTE LTE 81 201 T/2 21/3 T[/3 51/6 TT/6 TT OD 20 20 40 Žrt/6 111/6 + 1 Х p= (1 - a cos(0)) os(0)) 1 +](//img.homeworklib.com/questions/981490e0-9311-11eb-b709-7b502b443bb9.png?x-oss-process=image/resize,w_560)
![7:36 AM 1.6KB/s Vo LTE 4G JN LTE 81 101 1/2 2ít/B. TT/3) 51/6 TT/6 TT 0 20 10 10 20 71/6 1111/6 + 1 Х p= (1 – a cos(0) os(0))](//img.homeworklib.com/questions/98b78a20-9311-11eb-a68b-bd786754eff8.png?x-oss-process=image/resize,w_560)
according
to the Chegg guidelines I am bound to do only first four parts of
the question in case of multiple parts.
Including two sub parts of a I have already done 4 parts. If you
want any specific part then please mark that one
Sorry for inconvenience
4G 7:27 AM 0.1KB/s Vo LTE LTE 82 III 14 1/2 21/3 T1/3 51/6 f/6. TT 1 7/B 1111/6 41/3 51/3 + 1 w p=sin(n:0) Х 0 <O< 27 2 Х n = 2 -10 10 3 X 4 powered by desmos
4G 7:27 AM 0.OKB/s vo LTE LTE 82 III 1+ 1/2 21/3 T[/3 3 511/6 TT/ TT 1 711/6 1111/6 41/3 51/3 + 1 Х r = sin(n:0) <O< 7 0 2 Х n=3 -10 10 3 X 4 powered by desmos
4G 7:29 AM|0.OKB/s Vo LTE LTE 82 III 21/3 T/3 511/6 TT/ TT 1 AT/6 1116 41/3 51/3 + 1 Х r = sin(n·0)| 0 <o< 27 2 x n = 3 12 -10 10 3 Х 4 powered by desmos
4G 7:29 AM|1.5KB/S Vo LTE LTE 82 III 14 1/2 21/3 T1/3 51/6 f/6. TT 1 7/6 1 1/6 41/3 517/3 + 1 Х r = sin(n·0)| 0 <o< 27 2 n = 2 t2 -10 X • X 10 3 4 powered by desmos
4G 7:31 AM|1.7KB/s vo LTE LTE 82 4 2 1/2 21/3 13 51/6 TT/6 T 0 2 2 4 ZTt/6 111/6 41/3 51/3 + 1 N 7=1c sin(n :0) Х 0 <O< 26 2 n = 1 -10 10 X X . 3 c=1 -10 10 4. powered by desmos
4G 7:31 AM|0.0KB/s Vo LTE LTE 82 4 211/27 21/3 TT/3 5/6 TT/6 TT 0 2 2 4 ZTt/6 111/6 41/3 51/3 + 1 N 7=1c sin(n+ :0) Х 0 <O< 260 2 n = 3 -10 10 X X . 3 c=1 -10 10 4. powered by desmos
4G 7:33 AM | 10.2KB/S Vo LTE LTE 81 4 211/27 21/3 T1/3 5ſt/6 TT/6 2 2 4 71/6 1111/6 41/3 51/3 + 1 Х p=1+c sin(n.6 :0) 0 <o< 27 2 Х n = 3 -10 10 3 X c=2 -10 10 4. powered by desmos
4G 7:33 AM 0.0KB/s Vo LTE LTE 81 4 21/22 8/3 TT/3 51t/6 T/6 2 2 4 ZTt/6 111/6 41/3 Бт/3 + « 1 Х p=1+c sin(n.6 :0) 0 <o< 27 2 Х n = 3 -10 10 3 X c=4 -10 10 4. powered by desmos
4G 7:36 AM | 0.OKB/s Vo LTE LTE 81 III 101 T1/2 2/3 TT/3 51/6 TT/6 M 0 20 10 10 20 71/6 11T1/6 + 1 Х p= (1 - a cos(0)) os(0)) 1 + a cos 0 <O< 27 2 Х a = 1 -10 10 3 powered by desmos
4G 7:39 AM | 4.8KB/s Vo LTE LTE 81 201 T/2 21/3 T[/3 51/6 TT/6 TT OD 20 20 40 Žrt/6 111/6 + 1 Х p= (1 - a cos(0)) os(0)) 1 + a cos 0 <O< 27 2 Х a = 1.1 -10 10 3 powered by desmos
7:36 AM 1.6KB/s Vo LTE 4G JN LTE 81 101 1/2 2ít/B. TT/3) 51/6 TT/6 TT 0 20 10 10 20 71/6 1111/6 + 1 Х p= (1 – a cos(0) os(0)) 1 + a cos 0 <O< 27 2 Х a = 2 -10 10 3 powered by desmos
Oilas o= Suno n is odd. 2o. of loope equals for ton even for nis no. of loops equals to an. (b) In case of 8 - Süno) Both nis odd f even the no. of loops are an Q.2. gel + C Sivno • No. of loope inculares du There is in new loop due to Or Smallose los due to c. size of Small with increase in a constant n lineckily with with n loep inceland c for 0-3 r=1-a Caso It a coso Shape Changes as all Changer