All Answers 356007 Answers

sum(105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497, 504, 511, 518, 525, 532, 539, 546, 553, 560, 567, 574, 581, 588, 595, 602, 609, 616, 623, 630, 637, 644, 651, 658, 665, 672, 679, 686, 693, 700, 707, 714, 721, 728, 735, 742, 749, 756, 763, 770, 777, 784, 791, 798, 805, 812, 819, 826, 833, 840, 847, 854, 861, 868, 875, 882, 889, 896, 903, 910, 917, 924, 931, 938, 945, 952, 959, 966, 973, 980, 987, 994) =70,336

wow, I didn't see it that way until you broke it down. thanks

If 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(n*(n + 1)) = 99/100, what is n?

n = 99, so that:

If 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*(99 + 1)) = 99/100.

Use AP formula:
[659 - 17]=642 / 2 + 1=322 - number of terms.
[17 + 659] / 2 * 322 =108,836

There are 28 ways.

547 was arrived at as follows:

The LCM[12, 20, 36] = 180, then smallest N is:

180m + 7 =187, where m=0, 1, 2, 3......etc.

Since the number of students is between 500 and 600 students, then:

180 * 3 + 7 = 547 - Number of students in this school.

N mod 12 = 7

N mod 20 = 7

N mod 36 = 7

N = 547 - Number of students in the school.

(2016!-2015!) mod 2017 = 2015

Now I am not so sure, JL....the spring mass was started at equalibrium......so the mg of gravity was already being balanced by the spring.....THEN you ADDED  .25 x 50   N  of upward force which is what the weight experiences when you release it...I think this is correct.... the a will be 12.5 m/s^2 .

   The NET force will be upward due to the spring being further stretched......the spring at equalibrium has already balanced (i.e. cancelled out) the mg .....

  Did your teacher tell you the correct answer was 2.7 m/s^2 ? 

Your a guest. Indistinguishable from all other guests.

If you want to say something you need to say it properly.

Thanks :)

I just spent ages answering properly and now it is all gone.    

anyway this is not too hard.

Write the question with A,B and C properly.

There are only 3 circles.

A is totally inside B and C overlaps both of them.

See if you can do it now.

triangle>circle>rectangle hope this helps 

Both of the above answers are wrong!

112 mod 15 = 7 NOT 4

The smallest positive multiple of 7 that satisfies the 4 congruences is 364:

The LCM of 90m + 4, where m=0, 1, 2, 3.....etc.

Since the smallest multiples of 7 that satisfies the 4 congruences is:

4 x 90 + 4 = 364.

364 mod 6 = 4

364 mod 9 = 4

364 mod 15 = 4

364 mod 18 = 4

This is a different question.

Look at the first two statements!

I spent ages on this.

If you want to change a question like that please put the original one is as well so we can check it has been changed properly.

(1/3) / (5/1) =(1/3) x (1/5) =1 / 15

Really? The whole class?
10(3a-6)+14a
Distibute 10
30a - 60 + 14a
30a + 14a - 60
44a - 60. And that is it. If you want to solve for a, then:
44a - 60 = 0
44a = 60
a = 60 / 44 = 1+ 4/11

350 cubic inches =5735 cubic cm.
5735 - 5500 =235 cubic cm - difference between the two engines.

The smaller end of the range of possible values of   y - x   =   -5 - 2   =   -7

The bigger end of the range of possible value of     y - x    =   5 - -2   =   7

y - x  can be anything between  -7  and  7  by starting at one end and increasing or decreasing either  y  or  x

-7  <  y - x  <  7

for some reason, tihs had thrown me off :/

1ST. A

F  = ma

12.5 Newtons  =  12.5kg * m  / s^2

12.5 kg m / s^2  =  1.0 kg  *  a        [ divide out kg ]

12.5 m /s^2   =  a

Your second answer is correct.....look at your  first  one, again....