All Answers 356032 Answers

m = 3s, n = 9t, so mn = 27st will be divisible by 27.

Density=Mass/Volume

You have to find the volume of the block by immersing it in water. The amount of water displaced is the same as the volume of the block. The volume has to be in cubic cm or in milliliters.

The answer isn't 3.

PS, the recapcha thing validiator is so annoying! 

One cannot express a percent answer to this question; there are no numbers included in the word problem to work off of, only abstract ideas as to whether Sheila will attend a picnic. You need some given percentages.

Or use the 'minus' button on the calculator; put it directly to the left of the number you want to be negative. :)

(Which is right next to the 'equals' ( = ) button. 

Hi Vymoinh

Welcome to the web2.0calc forum    

Thanks Heureka :)

Hey guys, I need some help :( 

I have this question related about permutation and combination. Can you help me please? :< 

The 4-digits numbers are 5352, 5114, 5818,..., have 2 common characteristics. First, they start with the digit 5. Second, exactly 2 digits in each numbers are identical. How many such numbers are there?

11(the third digit can be (0,2,3,4,6,7,8,9)  

The third digit can be 1st 2nd or third so that it  8*3=24

There are 9 double digit possiblilities if you do not reuse 5 = 24*9=216 permutation.

Heureka's answer is different because he has allowed the 5 to be used once or twice.

So I would need to add all the ones with 2 fives.

9*8*3=216

216+216=432

Great, my answer is the same as Heureka's     

If you want more explanation then ask :)

2^2014=

1881097 3311373386 1250307391 6809514162 6221653231 0211821646 2569860015 3354426652 5952222275 0902380963 8726343818 6846451705 9008015759 7054780298 5500824593 3943444578 9451057421 5679639553 1284957564 7648951734 9153980520 0158634501 9877563824 2695491284 2154647000 7434593074 6360893274 2085748811 4585323380 6600051461 2761930651 1428634048 0136815578 9385479092 8639410329 1537818302 8780510521 3576072725 8123855923 2684216613 0002691129 1794057491 4593124841 2000918267 4703455439 3245932485 0518740263 7242832633 3186689598 7827870372 7546786415 2892763767 0384164665 1669597478 1074067878 4431329308 7433570387 2681455957 0815459944 3963660122 5697296384

\(\begin{array}{rl}&-\dfrac{1}{3}+(-\dfrac{3}{8})\\=&(-1)(\dfrac{1}{3}+\dfrac{3}{8})\\=&(-1)(\dfrac{8}{24}+\dfrac{9}{24})\\=&(-1)(\dfrac{17}{24})\\=&\color{red}-\dfrac{17}{24}\end{array}\)

\(\dfrac{10}{3}=3.\underbrace{333333333333333333333333333........}_\infty\)

Yes there is no end.

\(\dfrac{2\frac{1}{4}\text{ inches}}{\frac{1}{4}\text{ inches}}=\dfrac{60 \text{ inches}}{\text{what}}\)LOL xD

\((2\frac{1}{4}\text{ inches})(\text{what})=(60\text{ inches})(\frac{1}{4}\text{ inches})\)

\(\text{what} = \dfrac{15\text{ inches}^2}{2\frac{1}{4}\text{ inches}}=\color{red}6\frac{2}{3}\text{ inches}\)

2²⁰¹⁴ is so big that our calculator on this page can't calculate. And it's good to know that the unit digit of this number is 4.

\(2014\pmod4 \equiv 2\\ \therefore \text{Unit digit of }2^{2014} = \text{Unit digit of }2^2 = 4\)

Divide 52 by 50.

\(\dfrac{52 \text{ hours}}{50\text{ day}}=1.04 \text{ hours per day}\)

Hi Mellie :)

Find all real numbers a such that the roots of the polynomial x^3-6x^2+21x+a form an arithmetic progression and are not all real.

\(x^3-6x^2+21x+a\\ \mbox{Let the roots be }\alpha,\;\;\beta\;\;and\;\;\gamma\;\;\mbox{and let the common difference be }d\\ so\\ \alpha=\beta-d \quad and \quad \gamma=\beta+d\\~\\ \alpha+\beta+\gamma=6\\ 3\beta=6\\ \beta=2\\ \mbox{So the roots are }2-d,\;\;2,\;\;and\;\;2+d\\~\\ \alpha\beta+\alpha\gamma+\beta\gamma=21\\ 2(2-d)+(4-d^2)+2(2+d)=21\\ 4-2d+4-d^2+4+2d=21\\ -d^2+12=21\\ -d^2=9\\ d=\pm3i\\ \mbox{So the roots are }\;\;2-3i,\quad 2,\;\;and \quad 2+3i\\~\\ \alpha\beta\gamma=2(2-3i)(2+3i)=2(4+9)=26=-a\\so\\a=-26 \)

I have checked this result and it works.  I think it is the only solution.

\(u^2+33.405u-2360.582=0\)

\(u = {-33.405 \pm \sqrt{33.405^2-4(1)(-2360.582)} \over 2(1)}\) Quadratic Formula.

Use your calculator for the answer. Probably same answer as Guest #1

Solve for u:
u^2+33.405 u-2360.58 = 0

u^2+33.405 u-2360.58 = u^2+(6681 u)/200-1180291/500:
u^2+(6681 u)/200-1180291/500 = 0

Add 1180291/500 to both sides:
u^2+(6681 u)/200 = 1180291/500

Add 44635761/160000 to both sides:
u^2+(6681 u)/200+44635761/160000 = 422328881/160000

Write the left hand side as a square:
(u+6681/400)^2 = 422328881/160000

Take the square root of both sides:
u+6681/400 = sqrt(422328881)/400 or u+6681/400 = -sqrt(422328881)/400

Subtract 6681/400 from both sides:
u = sqrt(422328881)/400-6681/400 or u+6681/400 = -sqrt(422328881)/400

Subtract 6681/400 from both sides:
Answer: |u = sqrt(422328881)/400-6681/400    or     u = -6681/400-sqrt(422328881)/400

Solve for a:
25 = a^2/4+576/a^2

25 = a^2/4+576/a^2 is equivalent to a^2/4+576/a^2 = 25:
a^2/4+576/a^2 = 25

Bring a^2/4+576/a^2 together using the common denominator 4 a^2:
(a^4+2304)/(4 a^2) = 25

Multiply both sides by 4 a^2:
a^4+2304 = 100 a^2

Subtract 100 a^2 from both sides:
a^4-100 a^2+2304 = 0

Substitute x = a^2:
x^2-100 x+2304 = 0

The left hand side factors into a product with two terms:
(x-64) (x-36) = 0

Split into two equations:
x-64 = 0 or x-36 = 0

Add 64 to both sides:
x = 64 or x-36 = 0

Substitute back for x = a^2:
a^2 = 64 or x-36 = 0

Take the square root of both sides:
a = 8 or a = -8 or x-36 = 0

Add 36 to both sides:
a = 8 or a = -8 or x = 36

Substitute back for x = a^2:
a = 8 or a = -8 or a^2 = 36

Take the square root of both sides:
Answer: |a = 8    or    a = -8    or    a = 6    or    a = -6

52 / 50 =1 hour and about 2 1/2 minutes per day=1.04 hours per day.

2.25 / .25 =9

60 /9 =6 2/3

Therefore: 2 1/4:1/4 = 60:6 2/3

a^4 - 100a^2 - 2304  =  0      this can be factored as

(a^2 - 36) (a^2 - 64)   = 0

But how it can be factored? if       (-36)*(-64)=2304 but  you write -2304, and i need to get just one solution.

Thanks anyway for your help.

The 4-digits numbers are 5352, 5114, 5818,..., have 2 common characteristics. First, they start with the digit 5. Second, exactly 2 digits in each numbers are identical. How many such numbers are there?

\(\small{ \begin{array}{|rrrrrrrrr|} \hline 1. \quad5001 & 55. \quad5131 & 109. \quad5262 & 163. \quad5405 & 217. \quad5520 & 271. \quad5597 & 325. \quad5744 & 379. \quad5875 & \\ 2. \quad5002 & 56. \quad5133 & 110. \quad5265 & 164. \quad5411 & 218. \quad5521 & 272. \quad5598 & 326. \quad5745 & 380. \quad5877 & \\ 3. \quad5003 & 57. \quad5135 & 111. \quad5266 & 165. \quad5414 & 219. \quad5523 & 273. \quad5600 & 327. \quad5747 & 381. \quad5878 & \\ 4. \quad5004 & 58. \quad5141 & 112. \quad5272 & 166. \quad5415 & 220. \quad5524 & 274. \quad5605 & 328. \quad5750 & 382. \quad5880 & \\ 5. \quad5006 & 59. \quad5144 & 113. \quad5275 & 167. \quad5422 & 221. \quad5526 & 275. \quad5606 & 329. \quad5751 & 383. \quad5881 & \\ 6. \quad5007 & 60. \quad5145 & 114. \quad5277 & 168. \quad5424 & 222. \quad5527 & 276. \quad5611 & 330. \quad5752 & 384. \quad5882 & \\ 7. \quad5008 & 61. \quad5150 & 115. \quad5282 & 169. \quad5425 & 223. \quad5528 & 277. \quad5615 & 331. \quad5753 & 385. \quad5883 & \\ 8. \quad5009 & 62. \quad5152 & 116. \quad5285 & 170. \quad5433 & 224. \quad5529 & 278. \quad5616 & 332. \quad5754 & 386. \quad5884 & \\ 9. \quad5010 & 63. \quad5153 & 117. \quad5288 & 171. \quad5434 & 225. \quad5530 & 279. \quad5622 & 333. \quad5756 & 387. \quad5886 & \\ 10. \quad5011 & 64. \quad5154 & 118. \quad5292 & 172. \quad5435 & 226. \quad5531 & 280. \quad5625 & 334. \quad5758 & 388. \quad5887 & \\ 11. \quad5015 & 65. \quad5156 & 119. \quad5295 & 173. \quad5440 & 227. \quad5532 & 281. \quad5626 & 335. \quad5759 & 389. \quad5889 & \\ 12. \quad5020 & 66. \quad5157 & 120. \quad5299 & 174. \quad5441 & 228. \quad5534 & 282. \quad5633 & 336. \quad5765 & 390. \quad5895 & \\ 13. \quad5022 & 67. \quad5158 & 121. \quad5300 & 175. \quad5442 & 229. \quad5536 & 283. \quad5635 & 337. \quad5766 & 391. \quad5898 & \\ 14. \quad5025 & 68. \quad5159 & 122. \quad5303 & 176. \quad5443 & 230. \quad5537 & 284. \quad5636 & 338. \quad5767 & 392. \quad5899 & \\ 15. \quad5030 & 69. \quad5161 & 123. \quad5305 & 177. \quad5446 & 231. \quad5538 & 285. \quad5644 & 339. \quad5770 & 393. \quad5900 & \\ 16. \quad5033 & 70. \quad5165 & 124. \quad5311 & 178. \quad5447 & 232. \quad5539 & 286. \quad5645 & 340. \quad5771 & 394. \quad5905 & \\ 17. \quad5035 & 71. \quad5166 & 125. \quad5313 & 179. \quad5448 & 233. \quad5540 & 287. \quad5646 & 341. \quad5772 & 395. \quad5909 & \\ 18. \quad5040 & 72. \quad5171 & 126. \quad5315 & 180. \quad5449 & 234. \quad5541 & 288. \quad5650 & 342. \quad5773 & 396. \quad5911 & \\ 19. \quad5044 & 73. \quad5175 & 127. \quad5322 & 181. \quad5450 & 235. \quad5542 & 289. \quad5651 & 343. \quad5774 & 397. \quad5915 & \\ 20. \quad5045 & 74. \quad5177 & 128. \quad5323 & 182. \quad5451 & 236. \quad5543 & 290. \quad5652 & 344. \quad5776 & 398. \quad5919 & \\ 21. \quad5051 & 75. \quad5181 & 129. \quad5325 & 183. \quad5452 & 237. \quad5546 & 291. \quad5653 & 345. \quad5778 & 399. \quad5922 & \\ 22. \quad5052 & 76. \quad5185 & 130. \quad5330 & 184. \quad5453 & 238. \quad5547 & 292. \quad5654 & 346. \quad5779 & 400. \quad5925 & \\ 23. \quad5053 & 77. \quad5188 & 131. \quad5331 & 185. \quad5456 & 239. \quad5548 & 293. \quad5657 & 347. \quad5785 & 401. \quad5929 & \\ 24. \quad5054 & 78. \quad5191 & 132. \quad5332 & 186. \quad5457 & 240. \quad5549 & 294. \quad5658 & 348. \quad5787 & 402. \quad5933 & \\ 25. \quad5056 & 79. \quad5195 & 133. \quad5334 & 187. \quad5458 & 241. \quad5560 & 295. \quad5659 & 349. \quad5788 & 403. \quad5935 & \\ 26. \quad5057 & 80. \quad5199 & 134. \quad5336 & 188. \quad5459 & 242. \quad5561 & 296. \quad5660 & 350. \quad5795 & 404. \quad5939 & \\ 27. \quad5058 & 81. \quad5200 & 135. \quad5337 & 189. \quad5464 & 243. \quad5562 & 297. \quad5661 & 351. \quad5797 & 405. \quad5944 & \\ 28. \quad5059 & 82. \quad5202 & 136. \quad5338 & 190. \quad5465 & 244. \quad5563 & 298. \quad5662 & 352. \quad5799 & 406. \quad5945 & \\ 29. \quad5060 & 83. \quad5205 & 137. \quad5339 & 191. \quad5466 & 245. \quad5564 & 299. \quad5663 & 353. \quad5800 & 407. \quad5949 & \\ 30. \quad5065 & 84. \quad5211 & 138. \quad5343 & 192. \quad5474 & 246. \quad5567 & 300. \quad5664 & 354. \quad5805 & 408. \quad5950 & \\ 31. \quad5066 & 85. \quad5212 & 139. \quad5344 & 193. \quad5475 & 247. \quad5568 & 301. \quad5667 & 355. \quad5808 & 409. \quad5951 & \\ 32. \quad5070 & 86. \quad5215 & 140. \quad5345 & 194. \quad5477 & 248. \quad5569 & 302. \quad5668 & 356. \quad5811 & 410. \quad5952 & \\ 33. \quad5075 & 87. \quad5220 & 141. \quad5350 & 195. \quad5484 & 249. \quad5570 & 303. \quad5669 & 357. \quad5815 & 411. \quad5953 & \\ 34. \quad5077 & 88. \quad5221 & 142. \quad5351 & 196. \quad5485 & 250. \quad5571 & 304. \quad5675 & 358. \quad5818 & 412. \quad5954 & \\ 35. \quad5080 & 89. \quad5223 & 143. \quad5352 & 197. \quad5488 & 251. \quad5572 & 305. \quad5676 & 359. \quad5822 & 413. \quad5956 & \\ 36. \quad5085 & 90. \quad5224 & 144. \quad5354 & 198. \quad5494 & 252. \quad5573 & 306. \quad5677 & 360. \quad5825 & 414. \quad5957 & \\ 37. \quad5088 & 91. \quad5226 & 145. \quad5356 & 199. \quad5495 & 253. \quad5574 & 307. \quad5685 & 361. \quad5828 & 415. \quad5958 & \\ 38. \quad5090 & 92. \quad5227 & 146. \quad5357 & 200. \quad5499 & 254. \quad5576 & 308. \quad5686 & 362. \quad5833 & 416. \quad5965 & \\ 39. \quad5095 & 93. \quad5228 & 147. \quad5358 & 201. \quad5501 & 255. \quad5578 & 309. \quad5688 & 363. \quad5835 & 417. \quad5966 & \\ 40. \quad5099 & 94. \quad5229 & 148. \quad5359 & 202. \quad5502 & 256. \quad5579 & 310. \quad5695 & 364. \quad5838 & 418. \quad5969 & \\ 41. \quad5100 & 95. \quad5232 & 149. \quad5363 & 203. \quad5503 & 257. \quad5580 & 311. \quad5696 & 365. \quad5844 & 419. \quad5975 & \\ 42. \quad5101 & 96. \quad5233 & 150. \quad5365 & 204. \quad5504 & 258. \quad5581 & 312. \quad5699 & 366. \quad5845 & 420. \quad5977 & \\ 43. \quad5105 & 97. \quad5235 & 151. \quad5366 & 205. \quad5506 & 259. \quad5582 & 313. \quad5700 & 367. \quad5848 & 421. \quad5979 & \\ 44. \quad5110 & 98. \quad5242 & 152. \quad5373 & 206. \quad5507 & 260. \quad5583 & 314. \quad5705 & 368. \quad5850 & 422. \quad5985 & \\ 45. \quad5112 & 99. \quad5244 & 153. \quad5375 & 207. \quad5508 & 261. \quad5584 & 315. \quad5707 & 369. \quad5851 & 423. \quad5988 & \\ 46. \quad5113 & 100. \quad5245 & 154. \quad5377 & 208. \quad5509 & 262. \quad5586 & 316. \quad5711 & 370. \quad5852 & 424. \quad5989 & \\ 47. \quad5114 & 101. \quad5250 & 155. \quad5383 & 209. \quad5510 & 263. \quad5587 & 317. \quad5715 & 371. \quad5853 & 425. \quad5990 & \\ 48. \quad5116 & 102. \quad5251 & 156. \quad5385 & 210. \quad5512 & 264. \quad5589 & 318. \quad5717 & 372. \quad5854 & 426. \quad5991 & \\ 49. \quad5117 & 103. \quad5253 & 157. \quad5388 & 211. \quad5513 & 265. \quad5590 & 319. \quad5722 & 373. \quad5856 & 427. \quad5992 & \\ 50. \quad5118 & 104. \quad5254 & 158. \quad5393 & 212. \quad5514 & 266. \quad5591 & 320. \quad5725 & 374. \quad5857 & 428. \quad5993 & \\ 51. \quad5119 & 105. \quad5256 & 159. \quad5395 & 213. \quad5516 & 267. \quad5592 & 321. \quad5727 & 375. \quad5859 & 429. \quad5994 & \\ 52. \quad5121 & 106. \quad5257 & 160. \quad5399 & 214. \quad5517 & 268. \quad5593 & 322. \quad5733 & 376. \quad5865 & 430. \quad5996 & \\ 53. \quad5122 & 107. \quad5258 & 161. \quad5400 & 215. \quad5518 & 269. \quad5594 & 323. \quad5735 & 377. \quad5866 & 431. \quad5997 & \\ 54. \quad5125 & 108. \quad5259 & 162. \quad5404 & 216. \quad5519 & 270. \quad5596 & 324. \quad5737 & 378. \quad5868 & 432. \quad5998 & \\ \hline \end{array} }\)

There are 432 such numbers

Here's a quicker analytic method.

As a lead in, consider the first term, I x - 1 I.

If x is greater than 1, it would be calculated as  x - 1,

while if x is less than 1 it would calculated as 1 - x.

Same sort of thing for the others at x = 5 and x = 9.

(i) If x is less than 1, the equation becomes (1 - x) + (5 - x) = (9 - x), solution x = -3.

(ii) If x is less than 5 but greater than 1,  (x - 1) + (5 - x) = (9 - x), solution x = 5.

(iii) If x is greater than 9, (x - 1) + (x - 5) = (x - 9), solution x = -3. 

Note that 1111111=10^6+10^5+10^4+10^3+10^2+10+1

and 909091=10^6-10^5+10^4-10^3+10^2-10+1.

Compute the product of these two integers.

without mistakes:

geometric series:

\(\begin{array}{|rcll|} \hline a_n &=& a_1 \cdot r^{n-1} \\ s_n &=& a_1 \left( \frac{-1+r^n}{r-1} \right) \qquad \text{sum}\\ \hline \end{array}\)

geometric series 1:

\(\begin{array}{|rcll|} \hline && 10^6+10^5+10^4+10^3+10^2+10+1 \qquad | \qquad a_1 = 1 \qquad r = 10 \\\\ s_n &=& a_1 \left( \frac{-1+r^n}{r-1} \right) \\ s_n &=& -a_1 \left( \frac{-1+r^n}{1-r} \right) \\ s_n &=& -1\cdot \left( \frac{-1+10^n}{1-10} \right) \\ s_7 &=& -1\cdot \left( \frac{-1+10^7}{1-10} \right) \\ s_7 &=& -\left( \frac{-1+10^7}{1-10} \right) \\ \hline \end{array}\)

geometric series 2:

\(\begin{array}{|rcll|} \hline && 10^6-10^5+10^4-10^3+10^2-10+1 \qquad | \qquad a_1 = 1 \qquad r = -10 \\\\ S_n &=& a_1 \left( \frac{-1+r^n}{r-1} \right) \\ S_n &=& -a_1 \left( \frac{-1+r^n}{1-r} \right) \\ S_n &=& -1\cdot \left( \frac{-1+(-10)^n}{1-(-10)} \right) \\ S_7 &=& -1\cdot \left( \frac{-1+(-10)^7}{1+10} \right) \\ S_7 &=& -\left( \frac{-1-10^7}{1+10} \right) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && 1111111 \cdot 909091 \\ &=& (10^6+10^5+10^4+10^3+10^2+10+1) \cdot (10^6-10^5+10^4-10^3+10^2-10+1) \\ &=& s_7 \cdot S_7 \\ &=& -\left( \frac{-1+10^7}{1-10} \right) \cdot [-\left( \frac{-1-10^7}{1+10} \right)] \\ &=& \left[ \frac{(-1)^2-(10^7)^2}{1^2-(10)^2} \right] \\ &=& \frac{1-10^{14}}{1-10^2} \\ &=& \frac{10^{14}-1}{10^2-1} \\ &=& \frac{99999999999999}{99} \\ &=& 10101010101010 \\ \hline \end{array}\)

see:

Evaluate

\( \frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\).

\(\begin{array}{|rcll|} \hline && \frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7 \\ &=& \frac{2+1}{2}+\frac{4+1}{4}+\frac{8+1}{8}+\frac{16+1}{16}+\frac{32+1}{32}+\frac{64+1}{64}-7 \\ &=& \frac{2}{2}+ \frac{1}{2}+ \frac{4}{4} + \frac{1}{4}+ \frac{8}{8} + \frac{1}{8}+ \frac{16}{16} \frac{1}{16}+ \frac{32}{32}+ \frac{1}{32}+ \frac{64}{64} + \frac{1}{64}-7 \\ &=& 1+ \frac{1}{2}+ 1 + \frac{1}{4}+ 1 + \frac{1}{8}+ 1 + \frac{1}{16}+ 1+ \frac{1}{32}+ 1 + \frac{1}{64}-7 \\ &=& \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{16}+\frac{1}{32}+ \frac{1}{64}+6-7 \\ &=& \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{16}+\frac{1}{32}+ \frac{1}{64} -1 \\ &=& \frac{32}{64}+ \frac{16}{64}+ \frac{8}{64}+ \frac{4}{64}+\frac{2}{64}+ \frac{1}{64} -1 \\ &=& \frac{1+2+4+8+16+32}{64} -1 \\ &=& \frac{64-1}{64} -1 \\ &=& \frac{64}{64} - \frac{1}{64} -1 \\ &=& 1 - \frac{1}{64} -1 \\ &=& - \frac{1}{64}\\ \hline \end{array}\)