The matrix of a (7,9)was given.The address of the first byte
of a (1,1)matrix=1258.It takes 4 bytes to store the
number,then calculate the address of the last byte of a
(5,8) matrix??
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Answer Posted / manish panwar
1258-1261 1262-1265 1266-1269 1270-1273 1274-1277 1278-1281 1282-1285 1286-1289
1290-1293 1294-1297 1298-1301 1302-1305 1306-1309 1310-1313 1314-1317 1318-1321
1322-1325 1326-1329 1330-1333 1334-1337 1338-1341 1342-1345 1346-1349 1350-1353
1354-1357 1358-1361 1362-1365 1366-1369 1370-1373 1374-1377 1378-1381 1382-1385
1386-1389 1390-1393 1394-1397 1398-1401 1402-1405 1406-1409 1410-1413 1414-1417
(5,8) matrix
4 4 4 4 4 4 4 4 = (4*7)+3=31 1258+31 = 1289
4 4 4 4 4 4 4 4 = 4*8=32 1289+32 = 1321
4 4 4 4 4 4 4 4 = 4*8=32 1321+32 = 1353
4 4 4 4 4 4 4 4 = 4*8=32 1353+32 = 1385
4 4 4 4 4 4 4 4 = 4*8=32 1385+32 = 1417
(32*5)-1 = (160 -1)= 159
1258+159 = 1417
So answer 1410
| Is This Answer Correct ? | 2 Yes | 8 No |
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