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Okay, I am actually sorry,

I must concentrate on myself. 

Straight

I think Math and Physic forum, but actually a Math forum

Your rows and columns are: 9  x  14  ==126

And you are missing one row of "9s" ==990, 991, 992, 993, 994, 995, 996, 997, 998 ==9

126 + 9 ==135 - such integers.

I forgot-

990, 991, 992, 993, 994, 995, 996, 997, 998 (9) So there are actually 149 integers.

Also, it is not (19) I meant (10). Umm, can someone explain which answer is right and why?

Straight

There are:

(500, 505, 511, 515, 522, 525, 533, 535, 544, 545, 550, 551, 552, 553, 554, 556, 557, 558, 559, 565, 566, 575, 577, 585, 588, 595, 599, 600, 606, 611, 616, 622, 626, 633, 636, 644, 646, 655, 656, 660, 661, 662, 663, 664, 665, 667, 668, 669, 676, 677, 686, 688, 696, 699, 700, 707, 711, 717, 722, 727, 733, 737, 744, 747, 755, 757, 766, 767, 770, 771, 772, 773, 774, 775, 776, 778, 779, 787, 788, 797, 799, 800, 808, 811, 818, 822, 828, 833, 838, 844, 848, 855, 858, 866, 868, 877, 878, 880, 881, 882, 883, 884, 885, 886, 887, 889, 898, 899, 900, 909, 911, 919, 922, 929, 933, 939, 944, 949, 955, 959, 966, 969, 977, 979, 988, 989, 990, 991, 992, 993, 994, 995, 996, 997, 998)==135 such integers.

g ( f(x) ) =     2 * ( x+2)^3 - 5)   + 10

              =     2 (x+2)³                                     Yep !

Hello Guest,

Of the three-digit integers greater than $500$, how many have two digits that are equal to each other and the remaining digit different from the other two

500, 511, 522, 533, 544, 566, 577, 588, 599 (10)

600, 611, 622, 633, 644, 655, 677, 688, 699 (10)

700, 711, 722, 733 ,744 ,755 ,766, 788, 799 (10)

800, 811, 822, 833, 844, 855, 866, 877, 899 (10)

900, 911, 922, 933, 944, 955, 966, 977, 988 (10)

505, 515, 525, 535, 545, 565, 575, 585, 595 (10)

606, 616, 626, 636, 646, 656, 676, 686, 696 (10)

707, 717, 727, 737, 747, 757, 767, 787, 797 (10)

808, 818, 828, 838, 848, 858, 868, 878, 898 (10)

909, 919, 929, 939, 949, 959, 969, 979, 989 (10)

550, 551, 552, 553, 554, 556, 557, 558, 559 (10)

660, 661, 662, 663, 664, 665, 667, 668, 669 (10)

770, 771, 772, 773, 774, 775, 776, 778, 779 (19)

880, 881, 882, 883, 884, 885, 886, 887, 889 (10)

This gives 140 numbers, if it were so desired.

Hope that was helpful. ^^

Straight

2 sin A = 5 cos A.

sin a = 5/2 cos a

sin a / cos a = tan a = 5/2

Enter    10.1222222    then hit the ' = '

The figures that are convex poygons are B, C, D, F, G, and H.

Find the time for A and B together:

1 / (1 / A  + 1/B)  = T

    T = 60/11 = 5.55

they work for 3 hours ....this leaves   (4.55 - 3) / 5,55   job left = 0.4499625031247395967

Now all three work together and take two hours to complete the fraction of a job that is left:

.4499 /  ( 1/12 + 1/10 + 1/c)  =  2

c = 24 hours

If half of the 4/5 of 80 we subtract the 3/8 of 8/7 of 56, the result is:

1/2 ( 4/5 *80)   -   3/8 ( 8/7 * 56) =

4/10 * 80         -    24/56 * 56

   32               - 24

    = 8

y = 2 x^1/5  + 3      switch the x's and y's

x = 2 y^1/5  +3        now solve for y......you should be able to do that....

A similar way:

y = 2x ^1/5  + 3      solve for x =

                                   THEN switch the x's and y's

Hi Melody,

\( \sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}\)

\(\sqrt{245}=\sqrt{49*5}=\sqrt{49}*\sqrt{5}=7\sqrt{5}\)

\(\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}\)

\(\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}\)

Now Guest can do it from here.

Hope that was helpful.^^

Straight

18 I think

HCF(6,7,8) = 2

6*7*8/2 = 168  so it will be  mod 168

3*7*8  *  2*9  = 6*7*8*9

3*7*8  *    18  = 6*7*8*9

So i think there will be 18.

It would be good if you come back and comment on my answer.  

I've never studied modular arithmetic formally and I am guessing my way a bit.

Four people, 4 bracelets

One person gets exactly 2 so at least one person must get non.

2,1,1,0

4C2= 6 ways to choose the two

4 ways to chose the person who misses out

3 ways to chose the person to get 2,

2 ways to sort the rest

6*4*3*2 = 144 ways

2,2,0,0  

6 ways to pair the braclets, (normally you would halve this but I am thinking differently.)

4C2 ways to choose two people =6

6*6 = 36 ways

180 ways total, I think....

The circle centered at (2,-1) and with radius 4 intersects the circle centered at (2,8) and with radius sqrt(14) at two points A and B.

Find AB^2.

They don't intersect.

Either way.  Both hints help

I was thinking

\(ln(17x) = ln(17*x) \)

now use the rule I gave you to separate the two.  

How about   log (a/b) = log a - log b      Now you should understand....

= (x-1)⁵

F= ma

Now double m can be seen to  1/2 the  a   to 9 m/s²

thank you, but I still don't understand what to do.