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That means that there are 12 numbers that are not multiples of 5. Also, 8 integers that are multiples have to be even. 12+8 = 20 integers.

No wait, Guest's answer may be wrong if my calculations are correct. Since AT and AS trisect angle DAB, that means that angle SAD = 30 degrees and so triangle SAD is a 30-60-90 triangle. Since it's a 30-60-90 triangle, AD = sqrt3. Looking at TAD now, we get that it's a 30-60-90 triangle as well. That means that TD = sqrt3 * sqrt3 = 3. Adding CT and TD gives us that CD = 2sqrt3. Multiplying that by sqrt3, we get that the area of ABCD = 2sqrt3 * sqrt3 = 6. Hence, the area of ABCD is 6.

I am confused on how you got the 3 out of nowhere. Could you please explain more?

2. I don't see the image, could you please attach? Thanks!

1. You know that triangles EFC and BDE are similar by AA property. We also know that FC = 12-5 = 7. Since DE = 5 and FC = 7, you know that the similar ratio is 5/7. Since EF is a side of a square, it has a length of 5 as well. 5 * 5/7 = 25/7. Hence, BD has a length of 25/7.

This problem is simply asking how many numbers from 2 - 100 have 1 or more square factors. We first list out the multiples of the square factors and then eliminate common multiples.

4             9            We don't need to do 16.          25            We don't need to do 36.          49

8            18                                                           50                                                            98

12          27                                                           75

16                                                                               

20          45         

24          54         

28          63         

32                   

36          81         

40          90         

44          99         

48

52

56

60

64

68

72

76

80

84

88

92

96

100

25+9+3+2 = 39 integers that can be simplified.

We rewrite the equation as a + (a+1) + (a+2) + ... + (a+10583) = x^3. Adding them all up and factoring, we get (5292)(2a+10583) = x^3. We can factor 5292 as 3^3 * 14^2. That means that 2a+10583 = 14y^3, which has no solution because of the parity mismatch.

where do the points cross and how many solutions are there?

Solve for c:
(x y)/z = (a b)/c

(x y)/z = (a b)/c is equivalent to (a b)/c = (x y)/z:
(a b)/c = (x y)/z

Take the reciprocal of both sides:
c/(a b) = z/(x y)

Multiply both sides by a b:
 
c = (a b z)/(x y)

There are 16 such numbers:

558   576   588   666   678   696   756   768   786   798   858   876   888   966  978   996   Total =  16

GA how do you prove there are no rational solutions using fermat's last theorem? I don't understand

There is not much info there.

What formula do you want to use?

1. There is an acute angle theta such that cos(theta)=1/2+sin(theta). Find cos(2theta).

\(cos(\theta)=1/2+sin(\theta)\\ find\; \;\;cos(2\theta)\\~\ cos(2\theta)=cos^2\theta-\sin^2\theta\\ so\\ cos(2\theta)=(0.5+sin\theta)^2-\sin^2\theta\\ \)

In the interest of not promoting cheating. can you take it from there?

It would help if you included the address of the question that you want answered. 

The sum of the smallest two factors of an integer n is 6 and the sum of the two largest factor is 1122. Find n.

1 is a factor of every integer so the two smallest factors must be 1 and 5 which means that n is not divisable by 2 or by 3

The sum of the 2 largest factors is 1122.  So the number n must be smaller than 1122/2 = 561

So one is 561-x   and the other is 561+x

Maybe one is a multiple of 5           So maybe x=6t or 4t

x can't be 6t because 561 is divisable by 3 so 561*6t is also divisable by 3 which is not allowed.

What about x=4t

that would give us      561+4t    and   561-4t

I cannot see any obvious reason why this can't be true.  

I will try t values one at a time and see if I can find one that works

That one was easy, t =1 is a contender.

Well 565 does not have 2 or 3 as a factor and neither does 557 so they could be right.

557+565 = 1122

factor(565) = 5*113

factor(557) = 1+557

557*565 = 314705

So the number could be 314705

At this point in tme I am not convinced that this is the only possible answer though.

Hi Tom...riddle  

Wonderman is a pain BUT anybody who posts a question like that is in my veiw only interested in the quickest path to the answer.

i.e. they have the intention of cheating.

Sakath cannot even be bothered to present the question properly therefore I conclude he/she cannot be bothered to learn from any answer either.

Yes I spos, I just had not seen that before. 

2) Where's the picture?

Thanks SVS,

this is how I would attach this problem,

I had to look it up but I also have     2000 pounds in a US ton   .  

We have      \(\frac{x \;hours}{y \;pounds},\qquad \frac{2000\;pounds}{1\;ton}\qquad 1 ton,\qquad ?hours\\~\\ \)

I want   hours,   I want to cancel out all the other units.

\(\frac{x\;hours}{y\;pounds}\times \frac{2000\;pounds}{1ton}\times 1ton=\frac{2000x}{y}\;hours\)

Thanks Ginger,

That proof is so simple, I am glad you showed me.

4x - 6y = -12

     -6y = -4x - 12    | :(-6)

        y = 4/6x -12

Slope: 4/6 = 2/3

I do not know. I didn't work it out to the end. I left that to you.

You have been accused many times, from one person anyway, of cheating.

I do not mind helping you to learn and to develop solving techniques but I don't intend to give you final answers.

Thanks guest

p.a.  is a normal term in my part of the world.  It means per annum.

Does each function describe exponential growth or decay?