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   Elements  Containing                                     Number of subsets                              

    9 ,1    and  1 other possible element                2  =   sum of row 1  of Pascal's Triangle

    8, 2   and  2 other possible elements               4  =   sum of row 2

    7, 3  and 3 other possible elements                 8  =   sum of  row 3

    6, 4 and  4 other possible elements               16  =   sum of row 4 

                                                                           __

                                                                           30

Each set contains  two specific elements  and the the number of total  number of subsets with these = 

( 2) ^(possible  number of other elements) 

We know that \(ab = 6\), meaning the pairs of \((a,b)\) are (1,6), (2,3) (3,2), (6,1)

The coefficient of x will be \(b^2x + a^2x\)

Now, just try out the pairs, and find the least possible value.

The common ratio is \({\sqrt{10} \over \sqrt{2}} = \sqrt 5\)

The 7th term will be: \(\sqrt2 \times \sqrt5^6 = \color{brown}\boxed{125 \sqrt 2}\)

(a,h,k) = (3,-4,-12).

We have:

\(x = 2 (\mod 3)\)

\(x = 3( \mod 5)\)

\(x = 5 ( \mod 7)\)

Using the Chinese Remainder Theorem, we have \(x = 68 ( \mod 105)\), so the smallest value that works is \(\color{brown}\boxed{68}\)

Thank you!!!!!!!!

The above expression can be simplified to:

 \(\sqrt{\cfrac{1+\sqrt{\cfrac{1-\sqrt{\cfrac{1-\sqrt{\cfrac{1+\cfrac{\sqrt{3}}{2}}{2}}}{2}}}{2}}}{2}} = \sqrt{ \frac{1 + \cos\left(\frac{13\pi}{48} \right)}{2}} \)

The above value is the cosine of \(13\pi/96\)

So the answer to the question is: \(13\pi/96\)

-Vinculum

\( y = $\frac{1}{2}x^2$ - 9 \)

Call the point we are looking  for (c , c^2/2 - 9 )

Using the square of the  distance  formula  we have

D^2  =  ( c - 0)^2  + ( c/^2 / 2  - 9  - 0  )^2

D^2  = c^2  + c^4/4 - 9c^2 + 81

D^2  =  c^4 /4 - 8c^2 +  81           take the derivative

D^2 '   =   c^3 -16c            set to 0

c^3 - 16c  = 0

c ( c^2 - 16)  = 0

The second factor  set  = 0   is what we want

c^2  = 16

Two vaues of c  will  minimize the distance c = 4 or c  =-4......choose  c = 4

And y  = 4^2 / 2 - 9  =    -1

The point that minimizes the distance is  ( 4 , -1) 

a^2  =  4^2 + (-1)^2  =    17

\(8+x<4-3x-7<5x-4+11\)

\(8+x<4-3x-7\quad \mathrm{and}\quad \:4-3x-7<5x-4+11\)

\(x<-\frac{11}{4}\quad \mathrm{and}\quad \:x>-\frac{5}{4}\)

Since they are overlapping intervals:

No Solution

-Vinculum

You are currently 13 miles north and 8 miles east from your starting point.

Figuring it out using pythagorean theroem...: sqrt(13^2+8^2), answer = sqrt(233)

290               190                90 

190               200               100

 200              199               200

Start with the 290 and work your way to 100. From there, everything falls into place, although it is a hard problem for a 1st grader...

By trial and error find the smallest number to satisfy (i) and (ii) is 8.

Then go on by trialing numbers of 8 + 15n since 15 is the least common multiple of 3 and 5. 

You will find the answer to be 68 within just a few tries.

There should be a better and more official way by modulo arithmetic though so it would be great if someone can also add that to this thread.

Note that there is a \(4 \over 9\) chance to roll a 6,7, or 8. 

This means that the remaining \({5 \over 9}\)th of the time, you lose a dollar. 

The expected value is the average of the outcomes, and it is: \({{{(2 \times {4 \over 9})} + ({-1 \times {5 \over 9}})} \over 2 } = {3 \over 18}\)

So, your expected amount of winnings is \({3 \over 18 } \times 300 = \color{brown}\boxed{50}\)

x + sqrt (x - 2)  =  4

sqrt (x - 2)  = 4  - x             square both sides

x - 2 =  x^2 - 8x + 16         simplify as

x^2 - 9x +  18 =   0

(x - 3) ( x - 6)  =  0

x - 3 =  0                          x - 6 =  0

x =  3                               x =  6       (reject this solution)

x = 3  is the solution

Glue 2 of the boys together. There are \({4 \choose 2} = 6\) ways to do this. There are also 2 ways to order the 2 boys. 

Now, there are \(8!\) ways to order everyone (the 2 boys count as 1 person). 

This makes for \(8! \times 6 \times 2= \color{brown}\boxed{483840 }\) ways. 

There are more than one way to tackle this problem but I'll offer my best solution.

Total area of rectangle = 4*5 = 20

Then, subtract the three white parts that are not counted(2*4/2 = 4, 2*3/2 = 3, and 2*5/2 = 5)

Our final answer is 20 - 3 - 4 -5 = 8

I'd like to see someone answer this too.  

149,  228  [Both integers meet all your conditions]

9  +  8 ==17 - sum of all values of z

The function, g or k ? is linear, so its graph will be a straight line, in which case the gradient between x = 1 and x = 4  will be the same  as that between x = 4 and x = 13.

So,

\(\displaystyle \frac{b-13}{13-4}=\frac{13-a}{4-1}.\)

From that,

\(\displaystyle 3b-39=117-9a,\\ 9a+3b=156, \\ 3a + b = 52.\)

Alternatively, let the equation of the function be \(\displaystyle f(x) = Ax + B.\)

Substituting the co-ordinates of the three points, we have

\(\displaystyle a = A+B, \dots(1) \\13=4A+B, \dots(2) \\ b = 13A+B, \dots (3)\)

from which,

\(\displaystyle 3a+b = 3(A+B)+(13A+B)=16A+4B=4(4A+B)=52.\)

Good work!! 

if f(x) is perpendicular to g(x) then  a= -1/3

what is g(-1) ?

What is f( of your answer) ?

12 * 6 = 72 sqaure cm

3.14 * 2 * 2 = 12.56

72 - 12.56 -12.56 = 46.88

46.88 square cm

Wait... so \(x_1 = x_0 - x_ \text{-1}\), but the function isn't defined for \(n > 2\). 

The problem doesn't make sense.....

Add 1 to complete the square on x: \(x^2-2x +1 + y^2 = 25\)

Complete the square: \((x-1)^2 + y^2 = 25\)

The center occurs at \(\color{brown}\boxed{(1,0)}\), (x-y pair when the value of the circle equals 0). 

Also, note that the constant on the right-hand side of the equation equals the radius squared, so the radius is \(\sqrt {25} = \color{brown}\boxed 5\)