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Another answer is here

Querty,

Thanks for doing your reask properly and giving the address.

But please, next time, include the address in a proper link.

Dragon,

Thanks for answering but please answer on the original next time and just leave a note on the new ask to say that is what you have done. 

This is asking what is tan x approaching as x tends to pi/2 from the high side (from the right side going left)

It is approaching     minus infinity.

Can you see that?

let f(x)=y

\(y=\frac{23-3x}{6x}\\ y=\frac{23}{6x}-\frac{3x}{6x}\\ \)

That is a start, now first up you have to make x the subject.

Do that and then come back if you need more help.

How many natural numbers n have the property that the remainder of dividing 2003 by n is 23?

2003-23=1980

I need to find how many numbers bigger than 23 go into 1980

factor(1980 = ((2^2*3^2)*5)*11

1980 = 11* 5* 3* 3* 2* 2     

None of these factors on their own is bigger than 23

Look at pairs.

11*2 =22 which is too small, 11*3 and 11*5 are both fine.    So there is 2      (33  and 55)

5 any other than 11 is too small so no more here

Let me look at triples

11*2*2=44 is fine so possible multiples of 11 are fine    

11*2*3=66

11*2*5=110

11*3*3=99

11*3*5=165

5 are here

5*2*2=20 which is too small, 5*2*3=30, that is ok.   5*3*3=45,   5*3*2=30    are ok    so that is 2 more

3*3*2=18 which is too small, so no more here.

Now quads

11*2*2*3=121 that is ok, all quads with 11 are ok        

2*2*3*3=36 which is also ok so all quads are fine. 

2*2*3*3=36

2*2*3*5=60

2*2*3*11=132

2*2*5*11=220

2*3*3*5=90

2*3*3*11=198

2*3*5*11=330

3*3*5*11=495

8 of them

groups of 5 factors

leave out, 2,3,5 or 11       

4 of them.

6 factors

just 1 of them

2+5+2+8+4+1= 22

------------------------------

33

55

44

66

110

99

165

45

30

2*2*3*3=36

2*2*3*5=60

2*2*3*11=132

2*2*5*11=220

2*3*3*5=90

2*3*3*11=198

2*3*5*11=330

3*3*5*11=495

Plus 5 more

Thank you very much!

a=1; p=0;c=2003%a; if(c==23, goto4, goto6);p=p+1;printp,"-", a; a++;if(a<2000, goto2, 0)

OUTPUT:

      n

1 - 30
2 - 33
3 - 36
4 - 44
5 - 45
6 - 55
7 - 60
8 - 66
9 - 90
10 - 99
11 - 110
12 - 132
13 - 165
14 - 180
15 - 198
16 - 220
17 - 330
18 - 396
19 - 495
20 - 660
21 - 990
22 - 1980
 

Arc ACB is half the circle.  The whole circle is 360^o so ACB is 180^o   

Arc AC is given as 100^o which leaves 80^o for arc CB   

There's a relationship between the size of an angle subtended by an arc of a circle.  

When the angle's vertex lies on the circle the size of the angle is equal to half the arc.  

So angle CAB is 40^o   

BTW, when the angle's vertex lies on the center of the circle, the arc and the angle are the same amount of degrees. 

_.

We have x+30+x-30=180 which becomes 2x=180 so x=90 and x+30=y+20 so x+10=y and y=100 so z=-10 since 90-100=-10 so we have \(\boxed{z=-10}\)

Every musician in the band plays either the fife or the drum (but not both). The ratio of fifers to drummers is 7 : 1. If 2 drummers join the group and 6 fifers quit, the ratio of fifers to drummers would be 3 : 1. How many musicians were in the original ensemble?  

                                                                  F         7  

The ratio of fifers to drummers is 7 : 1.     ––   =   ––          Therefore, F  =  7D                                               (1)  

                                                                  D         1  

If 2 drummers join the group               D + 2          1

                                                           –––––   =   ––           Therefore 3D + 6  =  F – 6   thus   F  =  3D + 12    (2)

and 6 fifers quit                                   F – 6          3

Substitute 3D + 12 for F in equation (1)         3D + 12  =  7D   

                                                                                12  =  4D

                                                                                  3  =  D                           3 Drummers   

Substitute 3 for D in equation (1)                               F  =  7 • 3

                                                                                   F  =  21                        21 Fifers  

_.

The first die can be anything.  There's a 1/6 chance that the second die will match it. 

Another way of saying it is that there are 36 possible combinations.  Six of them are doubles.  6/36 = 1/6 

1,1   1,2   1,3   1,4   1,5   1,6  

2,1   2,2   2,3   2,4   2,5   2,6  

3,1   3,2   3,3   3,4   3,5   3,6  

4,1   4,2   4,3   4,4   4,5   4,6  

5,1   5,2   5,3   5,4   5,5   5,6  

6,1   6,2   6,3   6,4   6,5   6,6 

_.

@answer #1: Thanks for your answer, but I want to understand HOW to do this problem and I don't want just the answer. Could you please tell me how you did the problem? Thanks! 

Thanks Alan

See answer here:   https://web2.0calc.com/questions/can-anyone-solve-this_6

Dragan: Check this out. Does that look right??

arc XY = 170         arc VW = 110

arc VX + arc WY = 360 - ( 170+110 ) = 80

angle U = ( arc WY - arc VX ) /2 

30 = ( 70 - 10 ) / 2 

arc WY = 70º 

How do I factor and simplify this?​

Find the value of x

x = 30 

2x+2+5x+3=180

I will take a look at this:

a - 6 people can be seated, without restrictions, in 6! = 720 ways.

b - If one sits in a fixed seat, the other 5 can be seated in 5! = 120 ways.

c - 3 couples can be seated in: 3 x 3 = 9 ways.

9 =F + 4D,  -84 =F + 31D, solve for F, D

F =205 /9 - This is the first term.

D = - 31/9 - This is the common difference

23rd term =205/9 + ( - 31/9 * 22)

23rd term = -53

3 people had a burger                3 • 24  =   72  

1 person had a salad                  1 • 21  =   21  

all 4 had a drink                           4 •  8  =   32  

                                     total                      125   

There may be other combinations but when I found one, I quit looking. 

_.