in matlab y = -2pi to 2 pi (a) =tan(y-0.5) (b) = sin(y^2) (c) = cos(y^3) plot these in matlab in one frame
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- ɪɴ ᴊᴀᴠᴀ ᴡʜᴀᴛ ɪꜱ ɴᴇᴇᴅᴇᴅ ɪꜱ ᴀ ᴘʀᴏɢʀᴀᴍ ᴡʜɪᴄʜ ᴛᴀᴋᴇꜱ ᴀꜱ ɪɴᴘᴜᴛᴛᴡᴏ ꜱᴏʀᴛᴇᴅ ᴀʀʀᴀУꜱ ᴀɴᴅ ʀᴇᴛᴜʀɴꜱ ᴀ ɴᴇᴡ ᴀʀʀᴀУ ᴄᴏɴᴛᴀɪɴɪɴɢ ᴛʜᴇ ᴇʟᴇᴍᴇɴᴛꜱ ꜰᴏᴜɴᴅ ɪɴʙᴏᴛʜ ᴛʜᴇ ꜱᴏʀᴛᴇᴅ ᴀʀʀᴀУꜱ. ɪᴛ ɪꜱ ᴀʟʀɪɢʜᴛ ɪꜰ ᴛʜᴇ ɪɴᴘᴜᴛ ᴀʀʀᴀУꜱ ʜᴀᴠᴇ ᴅᴜᴘʟɪᴄᴀᴛᴇꜱ ʙᴜᴛ ᴛʜᴇʀᴇᴛᴜʀɴᴇᴅ ᴀʀʀᴀУ ꜱʜᴏᴜʟᴅ ʙᴇ ᴅᴜᴘʟɪᴄᴀᴛᴇ ꜰʀᴇᴇ ! ʙᴇʟᴏᴡ ɪꜱ ʜᴏᴡ ᴛʜᴇ ᴀʀʀᴀУꜱ ᴀʀᴇ ʀᴇᴘʀᴇꜱᴇɴᴛᴇᴅᴀʀʀᴀУ1[] = [1, 5, 6, 6, 9, 9, 9, 11, 11, 21]ʜᴇʀᴇ ʟᴇɴɢᴛʜ ᴏꜰ ᴀʀʀᴀУ1 ɪꜱ ᴍ. ᴀʀʀᴀУ2[] = [6, 6, 9, 11, 21, 21, 21]ʜᴇʀᴇ ʟᴇɴɢᴛʜ ᴏꜰ ᴀʀʀᴀУ2 ɪꜱ ɴ. ᴀʀʀᴀУ ᴛᴏ ʙᴇ ʀᴇᴛᴜʀɴᴇᴅ ᴡᴏᴜʟᴅ ʙᴇ:ᴀʀʀᴀУ[] = [6, 9, 11, 21] ( ᴅᴏ ɴᴏᴛ ᴜꜱᴇ ʟɪʙʀᴀʀУ ꜰᴜɴᴄᴛɪᴏɴꜱ ᴛᴏ ᴅᴏ ᴀɴУ ᴏꜰᴛʜᴇ ᴛᴀꜱᴋꜱ ʀᴇQᴜɪʀᴇᴅ ᴀʙᴏᴠᴇ) ᴡʜᴀᴛ ɪꜱ ᴛᴏ ʙᴇ ɪᴍᴘʟᴇᴍᴇɴᴛᴇᴅ ɪꜱ ᴀ ꜰᴜɴᴄᴛɪᴏɴ ɪɴ ꜱᴜᴄʜ ᴀ ᴡᴀУ ᴛʜᴀᴛ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ꜱᴏʟᴠᴇꜱᴛʜᴇ ᴘʀᴏʙʟᴇᴍ ᴡɪᴛʜ ᴏ(ᴍɴ) ᴛɪᴍᴇ ᴄᴏᴍᴘʟᴇxɪᴛУ. ᴏ(ᴍɴ) ɪꜱ ꜱᴀᴍᴇ ᴀꜱ ᴏ(ᴍ ∗ ɴ). Тʜɪꜱ ʙʀᴜᴛᴇ-ꜰᴏʀᴄᴇ ᴍᴇᴛʜᴏᴅ ꜱᴜɢɢᴇꜱᴛᴇᴅ ʜᴀꜱ ᴀ ɴᴀᴍᴇ ᴄᴀʟʟᴇᴅ ”ʟᴏᴏᴘ-ᴊᴏɪɴ” ᴡʜᴇʀᴇУᴏᴜ ʙᴀꜱɪᴄᴀʟʟУ ᴊᴜꜱᴛ ᴛʀᴀᴠᴇʀꜱᴇ ᴛʜʀᴏᴜɢʜ ᴛʜᴇ ᴇʟᴇᴍᴇɴᴛꜱ ᴏꜰ ᴏɴᴇ ᴀʀʀᴀУ ᴄᴏᴍᴘᴀʀɪɴɢɪᴛ ᴛᴏ ᴛʜᴇ ᴇʟᴇᴍᴇɴᴛꜱ ᴏꜰ ᴛʜᴇ ᴏᴛʜᴇʀ ᴀʀʀᴀУ ᴀꜱ ᴀ ᴄᴏᴍᴍᴇɴᴛ ɪɴ ᴛʜᴇ ᴄᴏᴅᴇ, ᴘʟᴇᴀꜱᴇ ᴡʀɪᴛᴇ ᴛʜᴇ ᴛᴇꜱᴛ-ᴄᴀꜱᴇꜱ ᴏɴ ʜᴏᴡ ᴏɴᴇ ᴡᴏᴜʟᴅᴛᴇꜱᴛ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ᴀꜱꜱᴜᴍᴘᴛɪᴏɴꜱ ᴀɴᴅ ʜᴇɴᴄᴇ ᴛʜᴇ ᴄᴏᴅᴇ.…Mar 18: 13 Jan 2:27 Sep 09 : 6 Aug 16: 192 Apr 5 : 44 Each line of the file lists the date of the reading and the number of broken-down busses that day. The dates are not in any particular order. The date is listed with the standard 3-letter month name abbreviation with the first letter capitalized, followed by a single space, then date in 1 or 2 digits (with a possible leading zero if date is less than 10). The date field is separated from the bus count with a colon, with any number of optional whitespaces before and after the colon. Write a function named filterDays that takes as parameters an input file name, an integer lower bound, and an integer upper bound. It should return a list of all days for which the broken bus count was between the lower and upper bounds (both inclusive). The items of the list should be tuples of the form (mm/dd', count), where mm and dd are 2-digit numeric values. For instance, with the sample input file above, a lower bound of 10, and an upper bound of…please solve within 30 minutes.
- 22+44/2=ʙᴇʟᴏᴡ ɪꜱ ʜᴏᴡ ᴛʜᴇ ꜰɪʟᴇɴᴀᴍᴇꜱ ᴀʀᴇ ʀᴇᴘʀᴇꜱᴇɴᴛᴇᴅᴀʀʀᴀʏ[] = [50, 11, 33, 21, 40, 50, 40, 40, 21]ʙᴇʟᴏᴡ ɪꜱ ᴛʜᴇ ᴇxᴘᴇᴄᴛᴇᴅ ᴏᴜᴛᴘᴜᴛ ᴀꜰᴛᴇʀ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴᴀʀʀᴀʏ[] = [11, 21, 33, 40, 50] ʏᴏᴜ ᴄᴀɴɴᴏᴛ ᴜꜱᴇ ʟɪʙʀᴀʀʏ ꜰᴜɴᴄᴛɪᴏɴꜱ ᴛᴏ ᴇɪᴛʜᴇʀꜱᴏʀᴛ ᴀɴᴅ ᴏʀ ᴘᴇʀꜰᴏʀᴍ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴏᴘᴇʀᴀᴛɪᴏɴ. ɪɴ ᴊᴀᴠᴀ ɪᴍᴘʟᴇᴍᴇɴᴛ ᴛʜᴇ ꜰᴜɴᴄᴛɪᴏɴ ɪɴ ꜱᴜᴄʜ ᴀ ᴡᴀʏ ᴛʜᴀᴛ ʏᴏᴜʀ ꜱᴏʟᴜᴛɪᴏɴ ꜱᴏʟᴠᴇꜱᴛʜᴇ ᴘʀᴏʙʟᴇᴍ ᴡɪᴛʜ ᴏ(ɴ ʟᴏɢ2(ɴ)) ᴛɪᴍᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ ᴏᴠᴇʀᴀʟʟ ʙᴜᴛ ꜱᴛᴀʏꜱ ᴡɪᴛʜɪɴ ᴛʜᴇ ʙᴏᴜɴᴅꜱ ᴏꜰ ᴏ(1) ꜱᴘᴀᴄᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ. ʜᴇʀᴇ, ɴ ɪꜱ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏꜰ ᴛʜᴇ ʟɪꜱᴛ ᴏꜰɪɴᴘᴜᴛ ɪɴᴛᴇɢᴇʀꜱ (ᴀʀʀᴀʏ). ɪ ʙᴇʟɪᴇᴠᴇ ᴛʜᴇ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ᴛʜᴀᴛ ᴄᴀɴ ʙᴇ ᴜꜱᴇᴅ ʜᴇʀᴇ ɪꜱ Qᴜɪᴄᴋ ꜱᴏʀᴛ. ᴘʟᴇᴀꜱᴇ ꜱᴛᴀᴛᴇ ᴀꜱ ᴄᴏᴅᴇ ᴄᴏᴍᴍᴇɴᴛ ᴡʜɪᴄʜ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ʏᴏᴜ ᴀʀᴇ ᴜꜱɪɴɢ, ꜱᴏʀᴛ ᴛʜᴇ ᴀʀʀᴀʏ ᴡɪᴛʜ ᴛʜᴀᴛ ᴀʟɢᴏʀɪᴛʜᴍ ᴀɴᴅ ꜱᴏʟᴠᴇ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘʀᴏʙʟᴇᴍ ᴛʜᴇʀᴇᴀꜰᴛᴇʀ. ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘᴀʀᴛ ᴏꜰ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ɪɴ ɪᴛꜱᴇʟꜰ ᴍᴜꜱᴛ ᴀᴅʜᴇʀᴇ ᴛᴏ ᴏ(ɴ) ᴛɪᴍᴇ ᴀɴᴅ ᴏ(1) ꜱᴘᴀᴄᴇ ʙᴏᴜɴᴅꜱ ᴄᴏᴍᴍᴇɴᴛ ᴇᴠᴇʀʏ ʟɪɴᴇ ᴏꜰ ᴄᴏᴅᴇᴘʟᴇᴀꜱᴇ ᴡʀɪᴛᴇ ᴛᴇꜱᴛ-ᴄᴀꜱᴇꜱ ᴏɴ ʜᴏᴡ ᴏɴᴇ ᴡᴏᴜʟᴅᴛᴇꜱᴛ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ᴀꜱꜱᴜᴍᴘᴛɪᴏɴꜱ ᴀɴᴅ ʜᴇɴᴄᴇ ᴛʜᴇ ᴄᴏᴅᴇ ᴇxᴀᴍᴘʟᴇ ᴏꜰ ᴄᴀꜱᴇꜱ ᴛᴏ ʙᴇ ᴛᴇꜱᴛᴇᴅ ꜰᴏʀ ᴀʀᴇ ʟɪᴋᴇ : ᴡʜᴀᴛ ɪꜰ ᴛʜᴇ ᴀʀʀᴀʏ ɪɴᴘᴜᴛ ᴡʜɪᴄʜɪꜱ…Question in image. Thank you for helping.
- Perform the following tasks by explaining the process of each one in detail: 1. Converting all numbers by follow between brackets: a. 487.2 from base 10 to [ bases 2,8, 16] b. 11010110 from base 2 to [ base 10 ] c. 456 from base 8 to [ base 10 ] d. 2B6 from base 16 to [ base 10 ] 2. Adding all numbers by follow between brackets: a. The binary numbers: [11010110 + 10000101] b. The octal numbers: [ 456 + 555 ] c. The hexadecimal numbers: [ 2B6 + 12C] 3. Subtracting all numbers by follow between brackets: a. The binary numbers: [ 11010110 - 10000101] b. The octal numbers: [ 555 – 456 ] c. The hexadecimal numbers: [ 2B6 - 12C]"""" HTML6.3 @l Write a fortran 90 program to move the elements of the first column &g to the last column and vice versa for the matrix B (/5,5/), then rotate &
- C 8 PDF assignment1.pdf File C:/Users/sebas/Downloads/HW1 Handout/handout/assignment1.pdf Q of 8 X New Tab x + col. + {B | An | (T) 1. A cell that has fewer than two live neighbors dies (because of loneliness) 2. A cell that has more than 3 live neighbors dies (because of over-crowding) 3. A cell that is dead and has exactly 3 live neighbors comes to life 4. All other cells maintain their state V Game of Life (Finally) So, for this step, change your life function so that it calls a new function named updateNextLife( oldB, newB ) in place of updateReversed, above. Then, implement the updateNextLife( oldB, newB ) function so that it sets each cell in the new data according to the updating rules based on the old generation, oldB: DI As suggested in updateReversed, always keep all of the outer-edge cells empty. This is simply a matter of limiting your loops to an appropriate range. However, it greatly simplifies the four update rules, above, because it means that you will only update the…8,9ʙᴇʟᴏᴡ ɪꜱ ʜᴏᴡ ᴛʜᴇ ꜰɪʟᴇɴᴀᴍᴇꜱ ᴀʀᴇ ʀᴇᴘʀᴇꜱᴇɴᴛᴇᴅᴀʀʀᴀʏ[] = [50, 11, 33, 21, 40, 50, 40, 40, 21]ʙᴇʟᴏᴡ ɪꜱ ᴛʜᴇ ᴇxᴘᴇᴄᴛᴇᴅ ᴏᴜᴛᴘᴜᴛ ᴀꜰᴛᴇʀ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴᴀʀʀᴀʏ[] = [11, 21, 33, 40, 50] ʏᴏᴜ ᴄᴀɴɴᴏᴛ ᴜꜱᴇ ʟɪʙʀᴀʀʏ ꜰᴜɴᴄᴛɪᴏɴꜱ ᴛᴏ ᴇɪᴛʜᴇʀꜱᴏʀᴛ ᴀɴᴅ ᴏʀ ᴘᴇʀꜰᴏʀᴍ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴏᴘᴇʀᴀᴛɪᴏɴ. ɪɴ ᴊᴀᴠᴀ ɪᴍᴘʟᴇᴍᴇɴᴛ ᴛʜᴇ ꜰᴜɴᴄᴛɪᴏɴ ɪɴ ꜱᴜᴄʜ ᴀ ᴡᴀʏ ᴛʜᴀᴛ ʏᴏᴜʀ ꜱᴏʟᴜᴛɪᴏɴ ꜱᴏʟᴠᴇꜱᴛʜᴇ ᴘʀᴏʙʟᴇᴍ ᴡɪᴛʜ ᴏ(ɴ ʟᴏɢ2(ɴ)) ᴛɪᴍᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ ᴏᴠᴇʀᴀʟʟ ʙᴜᴛ ꜱᴛᴀʏꜱ ᴡɪᴛʜɪɴ ᴛʜᴇ ʙᴏᴜɴᴅꜱ ᴏꜰ ᴏ(1) ꜱᴘᴀᴄᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ. ʜᴇʀᴇ, ɴ ɪꜱ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏꜰ ᴛʜᴇ ʟɪꜱᴛ ᴏꜰɪɴᴘᴜᴛ ɪɴᴛᴇɢᴇʀꜱ (ᴀʀʀᴀʏ). ɪ ʙᴇʟɪᴇᴠᴇ ᴛʜᴇ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ᴛʜᴀᴛ ᴄᴀɴ ʙᴇ ᴜꜱᴇᴅ ʜᴇʀᴇ ɪꜱ Qᴜɪᴄᴋ ꜱᴏʀᴛ. ᴘʟᴇᴀꜱᴇ ꜱᴛᴀᴛᴇ ᴀꜱ ᴄᴏᴅᴇ ᴄᴏᴍᴍᴇɴᴛ ᴡʜɪᴄʜ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ʏᴏᴜ ᴀʀᴇ ᴜꜱɪɴɢ, ꜱᴏʀᴛ ᴛʜᴇ ᴀʀʀᴀʏ ᴡɪᴛʜ ᴛʜᴀᴛ ᴀʟɢᴏʀɪᴛʜᴍ ᴀɴᴅ ꜱᴏʟᴠᴇ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘʀᴏʙʟᴇᴍ ᴛʜᴇʀᴇᴀꜰᴛᴇʀ. ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘᴀʀᴛ ᴏꜰ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ɪɴ ɪᴛꜱᴇʟꜰ ᴍᴜꜱᴛ ᴀᴅʜᴇʀᴇ ᴛᴏ ᴏ(ɴ) ᴛɪᴍᴇ ᴀɴᴅ ᴏ(1) ꜱᴘᴀᴄᴇ ʙᴏᴜɴᴅꜱ ᴄᴏᴍᴍᴇɴᴛ ᴇᴠᴇʀʏ ʟɪɴᴇ ᴏꜰ ᴄᴏᴅᴇᴘʟᴇᴀꜱᴇ ᴡʀɪᴛᴇ ᴛᴇꜱᴛ-ᴄᴀꜱᴇꜱ ᴏɴ ʜᴏᴡ ᴏɴᴇ ᴡᴏᴜʟᴅᴛᴇꜱᴛ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ᴀꜱꜱᴜᴍᴘᴛɪᴏɴꜱ ᴀɴᴅ ʜᴇɴᴄᴇ ᴛʜᴇ ᴄᴏᴅᴇ ᴇxᴀᴍᴘʟᴇ ᴏꜰ ᴄᴀꜱᴇꜱ ᴛᴏ ʙᴇ ᴛᴇꜱᴛᴇᴅ ꜰᴏʀ ᴀʀᴇ ʟɪᴋᴇ : ᴡʜᴀᴛ ɪꜰ ᴛʜᴇ ᴀʀʀᴀʏ ɪɴᴘᴜᴛ ᴡʜɪᴄʜɪꜱ…