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A number  written  as

a^15      with a being prime will have  16 divisors

But  2^15  is the smallest possible and it is   not a three  digit number

So.......

2^3 * 3 * 5 =  120 has 16 divisors                   

2 * 3 * 5 * 7 =  210  has 16 divisors                                

2^3 * 3^3  =  216  has 16 divisors                     

2 * 3^3   * 5  = 270 has 16 divisors                  

3 * 2^7  =  384    has 16 divisors                        

5 * 2^7 = 640  has 16 divisors                                     

2 * 3 * 5^3  =  750 has 16 divisors                    

2^7  * 7^1  =  896  has 16 divisors                     

{ I hope I haven't missed any  !!!   }

sumfor(n, 1, 10000, n^2009) mod 10^10==9,985,000,000 - these are the last 10 digits

We have the equation: \(x = {1 \over { 4 - x}}\).

Multiplying both sides by \(4 - x\) gives us: \(-x^2 + 4 x = 1\). 

Subtracting 1 from both sides gives us: \(-x^2 + 4x - 1 = 0 \)

Using the quadratic formula, we find \(x = \color{brown}\boxed{2 \pm \sqrt 3}\)

f(0)= 4 since z,e,r,o has four english alphabets, likewise f(1)& f(2)& f(6) will be equal to 6 and likewise it will go on till f(20) since the value of n is 0 to 20.

As we know domain means the input and that is the value of n and range mean the value of function. So comparing both we see that 0,1,2, 10 and numbers from 11-20 aren't included in the domain. Hence on the whole 14 numbers from 0-20 are not a part of range.

Let the intersection point directly above \(P\) as \(A\), and let the point that is to the left of \(P\) be \(B\)

The area of \(\triangle ABP = 12 \times 12 \div 2 = 72\) 

The area of the circular region \(ABP\) is \( {1 \over 4} \times 12^2 \times \pi = 36 \pi \)

Now, note that the area of the shaded region is \(2 (\text{circular region of ABP} - \triangle ABP) \)

Substituting what we know, we find that the area of the shaded region is \(\color{brown}\boxed{72 \pi - 144}\)

Note that \(8a^2b^2 = 8(ab)^2\)

We know that \(ab = 6\), meaning \(8a^2b^2 = \color{brown}\boxed{288}\)

Since UJ/UK = 7, let UJ = 7x and UK = x.  Also, JV/JK = 7, so let VJ = 7y and VK = y.  Then JK = 6y, so 8x = 6y so x = 3y/4.

UV = UK + KV = x + y = 7y/4.  W is the midpoint of UV, so WU = UV = 7y/8.  Then WJ = 6y + 7y/8 = 55y/8 and WK = WU - KU = 7y/8 - x = 7y/8 - 3y/4 = y/8

So WJ/WK = (55y/8)/(y//8) = 55.

Note that \(a^2 - b^2 = (a+b)(a-b)\).

Can you take it from here?

r = rate

d = distance

t = time

r = d/t

r = 710/70

r = 71/7

d = rt

d = 71/7*20

d = 1420/7

So the length of the jet is 1420/7 feet.

5^30 * 4^30   =

(5 * 4)^30  =

20^30

Note that

20^1  mod 7  =   6

20^2 mod 7  =   1

20^3 mod 7 =     6

20^4  mod 7  =  1

So......it appears that

20^(2n) mod 7  =  1

So

20^(2 * 15)  mod 7  =     20^30  mod 7   =  1

                     1

      _______________________    = 

     (x^2 + 3x - 4) (x^2 + 9x + 20)

                               1

         __________________________

           ( x + 4)(x - 1) (x + 5) ( x + 4)

The denominator cannot =   0.....   so

x = -5         x = -4          and x =  1            are not in the domain

                     A

            B      D       C

AD =  sqrt  [ AB^2  - [ (1/2)BC ] ^2 ]  =      sqrt [  25^2  -  9^2 ] =  sqrt [ 625 - 81 ]  =

sqrt [ 544 ]  ≈   23.32

There are four suits, 12 different possible consecutive triples, and 6 ways to order 3 things, so the probability is

4*12*6/C(52,3) = 228/52100 = 72/5525.

By brute force

The middle one  is either     5, 10,15,20,25,30,35  etc.

The  integers are   54 , 55  and 56

Sum  =   165

Your money   = Y

Your Brother's  money  =  B

So.....

Y - 5 =  B + 5     →      Y  =  B + 10        (1)  

Y + 14  =  2 (B - 14)         (2)

Sub (1)  into (2)   for Y

(B + 10) + 14  = 2(B - 14)

B + 24  =  2B  - 28

24 + 28   =  B

52  = B        { Your brother has $52  }

Call the capacity, C

So......

(1/6)C + 2   =   (5/12)C

2 =   (5/12)C  - (1/6)C

2=  (5/12)C  - (2/12)C

2 =  (3/12)C

2 = (1/4)C

2 (4/1)  = C   =     8  gallons

(3x + 12)(3x + 4) = (4x^2 + 8x + 18x^2)/(2x)       

9x^2 + 48x + 48  =  ( 2x)  ( 2x + 4 + 9x)  / (2x)     .....  assuming that x  does not  = 0 

9x^2 + 48x + 48  =  2x + 4 + 9x

9x^2 +  48x + 48  = 11x  + 4             rearrange  as

9x^2 + 37x + 44   =  0

Using the Q Formula

              -37 ± √ [ 37^2  - 4 *9 * 44]              -37 ± √[ -215]                -37 ± √[215] i

x  =     _______________________  =     _____________ =      _____________

                           2 * 9                                        18                                 18

What is a+b?

Hello Guest!

\(g(x)=2/f(x)\\ f(x)=3x^2+4x+2\\ \frac{df(x)}{dx}=6x+4=0\\ x_{max}=-\dfrac{2}{3}\\ {\color{blue}g(x)_{max}=}2/(3\cdot \frac{4}{9}-\frac{8}{3}+2)=\color{blue}3\)

\(R\in \mathbb R\)  |  0 < R \(\leq\) 3\(\)

\(\color{blue}a + b = 0 + 3 = 3\)

  !

n + (n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)==14 * (n+6), solve for n

n ==- 9

-9, - 8, - 7, - 6, - 5, - 4, - 3== - 42

- 3  x  14 ==- 42

B==Boys,      G==Girls,    

B  -  G ==90...................(1)
2/5B - 3/7G==25............(2), solve for B, G

B ==475
G==385

B  +  G ==475  +  385 ==860 children were invited.

How to get 512.51

cos A = 6/19.

BD = sqrt(2).

By Ptolemy, BD^2 = 63/5.

Here is the diagram: