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If you add the digits of each multiple of 9, you get a multiple of 9, eg:36=3 + 6 = 9783=7 + 8 + 3 = 18One digit is missing from each of the multiples of 9 below. Enter the missing digit for each number.71_2_201_1 010_000 000?
Yusra Mahmood
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Yusra Mahmood
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Swap 2 variable without a 3rd variable in c program?
This can be done using the bitwise exclusive-or(XOR) operator. int X = 16 // Binary 010000 int Y = 32 // Binary 100000 X = X ^ Y // X = 110000 Y = X ^ Y // Y = 010000 X = X ^ Y // X = 100000 This is good to know but is not widely practiced, as it is far simpler to use a temporary variable, and less prone to bugs. int temp = X; X = Y; Y = temp;
What is the password for Level 10-8 on the Starfy Nintendo game?
Simple... THERE ISN'T ONE! The numbers on the door represent the time you must get in each level to open the door. For example (This is only an example and is not the actual number I would believe): 10-1 010000 would mean you have to complete level 10-1 in under 1 minute. Colon are placed in correct spots to help: 01:00:00. See the relativity?
How many centimeters are in decameter?
AnswerI believe you are asking how many centimeters are in a decameter.The simple answer is 1000.Kilo Hecto Deca Meter Deci Centi Milli.010000 0.1000 1.0000 10.000 100.00 1000.0 10000You simply move the decimal point 4 places to the right. You have 1 decameter or in the illustration above 1.0000 decameters. Centimeter is 4 places to the right of decameter so move the decimal point the many places and you have 1000.0 centimeters.
How can you rewrite a perfect square as a exponent?
Find its square root and raise it in the second power.64 = 8281 = 92841 = ?2 you need to work a little bit here1. base would be in two digits (odd number with 3 digits, (3+1)/2)2. the first digit would be 2 (the square of 2 is less than 8), so the base would be between 20 and 30.3. the second digit would be 9 (the square of 9 ends in 1)4. the answer is 29, 841 = 292529 = ?21. base would be in two digits (odd number with 3 digits, (3+1)/2)2. the first digit would be 2 (the square of 2 is less than 5), so the base would be between 20 and 30.3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 23 or 27?4. 529 - 2232 = 529 - 409 = 120; it is 3, because 2(2*3)=12. So 529 = 2322809 = ?21. base would be in two digits (even number with 4 digits, 4/2)2. the first digit would be 5 (the square of 5 is less than 28,), so the base would be between 50 and 60.3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 53 or 57?4. 2809 - 5232 = 2809 - 2509 = 300; it is 3, because 2(5*3)=30. So 2809 = 5327056 = ?21. base would be in two digits (even number with 4 digits, 4/2)2. the first digit would be 8 (the square of 8 is less than 70,), so the base would be between 80 and 90.3. the second digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 84 or 86?4. 7056 - 8242 = 7056 - 6416 = 640; it is 4, because 2(8*4)=64. So 7056 = 8427396 = ?21. base would be in two digits (even number with 4 digits, 4/2)2. the first digit would be 8 (the square of 8 is less than 73), so the base would be between 80 and 90.3. the second digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 84 or 86?4. 7396 - 8262 = 7396 - 6436 = 960; it is 6, because 2(8*6)=96. So 7396(if you test 4 first, you'll see that it doesn't work, so it will be 6)55696 = ?21. base would be in three digits (odd number with 5 digits, (5+1)/2)2. the first digit would be 2 (the square of 2 is less than 5), so the base would be between 200 and 300.3. the last digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 2_4 or 2_6?4. let's work a little bit here:2_4; 22_42 = 4001640016+ ... = 4569600080 (101*2*4*_, we need to find the middle digit)01600 (102*2*2*4)04000 (103*2*2*_)55696 - 45696 = 10000400160008001600040001000000080 doesn't change, this tell us the middle digit can be 1 or 6. But one is too small and 6 is too big (2*2*1= 4, 2*2*6=32). So the last digit is not 4, it is 6. Let's find the middle digit.2_6; 22_62 = 4003640036+ ... = 4655600120 (101*2*6*_, we need to find the middle digit)02400 (102*2*2*6)04000 (103*2*2*_)55696 - 46556 = 9140400360012002400040000914000120 becomes 00160, and tell us the middle digit can be 3 or 8, but 8 is too big (2*2*8=32), so it is 3Thus, 55696 = 2362.263169 = ?21. base would be in three digits (even number with 6 digits, 6/2)2. the first digit would be 5 (the square of 5 is less than 26), so the base would be between 500 and 600.3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 5_3 or 5_7?4. let's work a little bit here:5_3; 52_32 = 250009250009+ ... = 263069000060 (101*2*3*_, we need to find the middle digit)003000 (102*2*5*3)010000 (103*2*5*_)263169 - 263069 = 10025000900006000300001000000010000060 doesn't change so the only possibility is when the middle digit is 1.Thus, 2631169 = 5132
How can you find out what kind of graphics card you have if the 'dxdiag' command pulls up something that doesn't have a video tab?
The debug command is usually helpful in identifying the video card. To start debug: Click on "Start" then click "Run" Type in "debug" then click "OK" You will be presented with a "-" prompt. Type in "d c000:0000" then press enter. You will see something like this: C000:000055 AA 78 EB 4B 37 34 30-30 E9 4C 19 77 CC 56 49U.x.K7400.L.w.VIC000:001044 45 4F 20 0D 00 00 00-38 01 60 0D 00 00 49 42DEO ....8.`...IBC000:00204D 20 56 47 41 20 43 6F-6D 70 61 74 69 62 6C 65M VGA CompatibleC000:003001 00 00 00 B0 10 2E AC-30 38 2F 31 33 2F 30 33........08/13/03C000:004000 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00................C000:0050E9 BC E2 00 00 00 00 00-3F D0 30 7E 80 20 40 81........?.0~. @.C000:006000 00 00 00 10 00 00 80-22 00 A5 8B E9 B7 C9 E9........".......C000:0070BE C9 50 4D 49 44 6C 00-6F 00 00 00 00 A0 00 B0..PMIDl.o.......The information we want to be looking at is in the third column. If no valuable information is found, type "d c000:0070" then press enter. C000:0070BE C9 50 4D 49 44 6C 00-6F 00 00 00 00 A0 00 B0..PMIDl.o.......C000:008000 B8 00 C0 00 33 0A F6-00 00 00 00 00 00 00 00.....3..........C000:0090FF 7F 4E 56 00 05 27 B2-1A 78 27 20 34 04 00 00..NV..'..x' 4...C000:00A000 00 00 00 00 00 FF 06-00 00 00 00 00 00 50 01..............P.C000:00B050 00 07 4D A0 19 14 3D-6E 3D 20 44 D5 43 E9 43P..M...=n= D.C.CC000:00C010 02 33 02 49 6E 00 00-00 00 00 00 00 00 06 C8..3.In..........C000:00D0A7 5A A8 60 AE 0A 00 30-57 05 00 82 6E 40 6F 48.Z.`...0W...n@oHC000:00E06F 68 6F 98 6F 98 6F 40-6F 70 DB 32 DC 00 00 00oho.o.o@op.2....In this case, there is still no information which identifies the video card, so we want to type "d c000:00e0" then press enter. C000:00E06F 68 6F 98 6F 98 6F 40-6F 70 DB 32 DC 00 00 00oho.o.o@op.2....C000:00F000 00 00 00 00 C4 DC C5-DC 08 A9 D8 A8 00 00 00................C000:010000 00 00 09 E5 AE D7 F8-A8 E8 A8 00 00 6C 6E 4F.............lnOC000:01106E 00 00 00 00 00 00 00-00 00 00 FF FF 02 00 00n...............C000:012048 E4 4F E4 55 E4 B2 6F-C6 E5 00 00 28 23 18 01H.O.U..o....(#..C000:013065 6E 64 20 62 6D 70 00-50 43 49 52 DE 10 22 03end bmp.PCIR..".C000:014000 00 18 00 00 00 00 03-78 00 01 00 00 80 00 00........x.......C000:015047 65 46 6F 72 63 65 20-46 58 20 35 32 30 30 20GeForce FX 5200This time, we can see information which identifies our video card. In this case, a GeForce FX 5200. If the video card still can't be identified by this step, try the following commands, pressing enter after each one. d c000:0150d c000:01c0d c000:0230If after these 3 additional commands, the video card isn't identified, the case will need to be opened to physically inspect the video card.
