Simplify the following expression,3 (-8 x –6 × 3) |4 – 12| – (6 –4)*]115634441148-3468348

First Solve the Parenthesis Bracket = (-8 * -6 * 3) = 144 Then Solve The |4 - 12| = |-8| = 8 The…

Answered: Simplify the following expression, (-8…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Perform the addition. 11,0112+11,0112
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H - For the IEEE 754 single-precision floating point, write the hexadecimal representation for the following decimal values: (i)–1.0 (ii)– 0.0 (iii)256.015625
1): The prime factorization of 198 is 198 = 2 · 32 · 11 A): 2 · 3.3· 11 B): 2.33.3 C): 1.198 D): 18.11 2): Some numbers and their prime factorizations are given below. 140 = 22 5 · 7 175 = 52 7 532 = 22 7 · 19 648 = 23 34 1078 = 2 · 72 11 1083 = 3 · 192 15435 = 32 5 · 73 1078 = 2 · 72 11 25480 = 23 5 · 72·13 Using these prime factorizations the gcd(532, 15435) = A): 7 B): 52 C): 97 D): 77 3): Applying the Euclidean Algorithm and the Extended Euclidean Algorithm. the gcd(81,60) , and expression of the gcd as a linear combination of the two numbers = A): gcd=3, combination = 3.81-4.60 B): gcd=3, combination = 21-18 C): gcd=12, combination = 81-69 D): gcd=9, combinaton = 81-60-12 4): using arithmetic in the ring of integers mod 9, 5×8inZ9= A): 4 B): 36 C): 2 D): 0
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…
Compute the following modular exponents.5^5mod13 = 7^9mod8 = 9^10mod11 =
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…

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Simplify the following expression, (-8 x –6 × 3) |4 – 12| (6 – 4)*| 1156 3444 1148 -3468