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Here's what Wolframalpha finds  

https://www.wolframalpha.com/input/?i=%283x%5E2-12x+%2B+16%29+%282y%5E2+%2B+6y%2B+9%29++%3D+18

The largest positive integer less than 100 with only 2 positive divisors = 97 [ 1 and 97]

The smallest integer with exactly 3 positive divisors = 4 [1,  2  and 4]

[97  +  4] = 101

How many even perfect square factors does 2^4*5^5 have?

P^2=( 2, 4, 10, 20,  50, 100)>>Total = 6

I'm assuming that by prime-looking you mean any number that is NOT divisible by ALL three numbers [2, 3 and 5]. If that is correct, I can only think of 3 such numbers:

The numbers are: 49,  77   and  91

Write the sum as a produkt.

\(sin(x)-sin(2x)\)

Hello Guest!

\(sin(x)-sin(2x)\\ =sin(x)-2sin(x)cos(x)\\ \color{blue}=sin(x)(1-2cos(x))\)

  !

lcm(1! + 2! + 3!, 2! + 3! + 4!, 3! + 4! + 5!, 4! + 5! + 6!, 5! + 6! + 7!, 6! + 7! + 8!, 7! + 8! + 9!, 8! + 9! + 10!)

=4,572,288,000=2,286,144,000  x  2

(b, a) =[2,286,144,000,  2]

[b + a] =[2,286,144,000  +  2] =2,286,144,002

Area of   ABC =  (1/2) (AB) (BC) sin (BAC)     (1)

Area of AXY  = (1/2) (AX) ( AY) sin (BAC)      (2)

Dividing  ( 2)  by (1)   we will  "cancel"  (1/2)  and sin (BAC)  ...so we  get

        AX  * AY                 4 * 6             24            12   

       ________  =         ______   =     ___   =    ____

         AB  * BC               7 * 10            70            35

Write the expression in terms of sine only

\(-\sqrt{3}\ sin(x)+cos(x)\)

Hello Guest!

\(-\sqrt{3}\ sin(x)+cos(x)\\ \color{blue}=\sqrt{3}\ sin(x)\pm \sqrt{1-sin^2(x)}\)

  !

Constant term   

C(6,3) (6x)^3  ( 1 / (3x) )^3  =

20 * (6)^3 * x^3  *  ( 1 / 27)  * ( 1/x^3)  =

20 * 6^3 / 27  =

20 * 216 / 27  =

20  *  8  = 

160

Verified here    :   https://www.wolframalpha.com/input/?i=%286x+%2B+1%2F+%283x%29+%29%5E6

Wavelength = Speed of light / Frequency

                       =[299,792,458 m/s] / [5.45 x 10^14 Hz]

                       =5.5 x 10^-7 meters.

(x + 1)^3  =  x^3  + 3x^2  + 3x  + 1

All  we  require is that  some  q(x)  can be  found such that  the x^3  and 3x^2  terms in the first polynomial can be canceled

Let  q (x)   be     (-x - 3)

So notice that

x^2   * q(x)  =   x^2  (-x - 3)   =  -x^3 -3x^2

So

x^3  + 3x^2  + 3x  + 1    +   (-x^3 - 3x^2)  =

3x  + 1

And  this is a polynomial  of  less than degree 2

(Note that  the choice of q(x)  isn't unique.... there are many other possibilities)

Not as difficult as it  seems.....we can use an identity to simplify  the given expression as  :

cos ( 2 θ  -  θ)  =  1/2

cos   θ  = 1/2

And  this happens   at    60°  =   pi/3  rads  =  θ

And at    300°  =  5pi/3 rads =  θ

 θ =  pi/3 , 5pi/3

LCM of 1152 and 315 = 8!
GCD of 1152 and 315 = 9
LCM of 315 and 1152 = 8!
GCD of 315 and 1152 = 9

1 + cot^2 (x) = [4 cos^2 (x) / (sin^2 (2x))]

Hello Guest!

\(1 + cot^2 (x) =\frac{ 4 cos^2 (x)} { sin^2 (2x)}\)

\(1 + cot^2 (x) =\frac{ 4 cos^2 (x)} { 4sin^2(x)cos^2(x)} \)

\(1 + cot^2 (x) =\frac{1}{sin^2(x)}\\ sin^2(x)=\frac{1}{1+cot^2(x)}\)  True!

  !

sin ( arccos (2/3) - arctan (1/3) )

Let  arccos (2/3)  = A  ...so ...    cos A  = 2/3    ......so    sin A  = sqrt (3^3 - 2^2) / 3  =  sqrt (5) / 3  

 and  arctan (1/3) ...   so ..  tan B =  1/3  ......so     sin B =   1/ sqrt ( 1^2 + 3^2)  =  1 / sqrt (10)    and cos B = 3 / sqrt (10)

So we have  

sin ( A  -  B  )  =   sin A cos B    - sin B cos A  

sin ( A - B) =  sqrt (5) / 3  *  3 /sqrt (10)   -  1/sqrt (10) * (2/3) 

sin (A - B)  = sqrt (5) / sqrt (10)  - 2 / (3 sqrt (10)

sin (A - B) =    1/sqrt (2)  - 2sqrt (10)  / 30

sin ( A - B)  =  1/sqrt (2)  - sqrt (10) / 15

sin (A - B)  =  sqrt (2) / 2  -  sqrt (10) /15

sin (A - B)  =   15sqrt (2)  - 2sqrt (10)

                     ___________________    

                              30

If tan (x)  =-4/3  and x is in Q2....then y is positive, x is negative and  r   =  sqrt   ( 3^2  + 4^2)  = sqrt  (25)   = 5

sin x  =  4/5        cos x =  -3/5

sin (2x)   =2 sin (x) cos (x)  =  2  ( 4/5) (-3/5)  =  -24/25

cos (2x)   =  cos^2(x)  - sin^2(x)  = (-3/5)^2 - (4/5)^2  =  9/25 - 16/25  =  -7/25

tan (2x)  =     sin(2x)   / cos (2x)  =    (-24/25) / ( -7/25)  =  -24/-7  =  24/7

Note....probability of picking any  red  = 10/15  = 2/3

And probability of picking any green = (5/15) = (1/3)

P(4 green peppers) =  C(8,4)  ( 2/3)^4 (1/3)^4  =  1120/6561

P(5 green peppers)  = C(8, 5)  = (2/3)^3 (1/3)^5  = 560/6561

Adding these   ( 1120  + 560) / 6561  = 1680 / 6561   =  560 / 2187

True.....

arcsec (0.5)   means that  the secant of some angle  = 0.5......but this is impossible because the secant only has a range of    (-inf , -1]  U [1, inf)

Thanks! 

Note  that we just need  two things that multiply  to  7  = (7 and 1)

[We could also use -7 and -1, but its not common to use lead negatives in  factoring.....the results would be the same even if we did ]

And we need two  things  that multiply to   -11 = ( 1 and -11)  or  ( 11 and -1))

So....we  would have

(7x + 1) ( x - 11)

(7x - 1)  (x + 11)

(7x + 11) ( x -1)

(7x - 11) ( x + 1)

Two positive integers m and n are chosen such that m is the largest positive integer less than 100 with only two positive divisors and n is the smallest integer with exactly three positive divisors. What is m + n?  

If the number is to have only two divisors,

then at least one of the divisors has to be prime.  

So how about taking the largest prime less than 50,     

and doubling it, so that would be 47 x 2, thus m = 94  

If 1 is qualified to be considered a divisor, then m = 97 but I don't like that. 

The other number, smallest with three divisors, I just used the  

three smallest factors, so, that would be 2 x 3 x 5, thus n = 30 

Final answer, m + n = 124  

If you wanted to qualify 1 as a divisor, you can find n the same way.  

Stop trying to cheat on AoPS homework.

ugh next time im using my account. please don't pretend to be another person.