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Those aren't appropriate answers because they weren't part of the multiple choice. You must choose from the choices provided. 

At the top of the pyramid, I believe 9⁰ should equal 1.  

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I don't know anything about Sufism, but my guess at the answer is

Sufis believe that Christian people do not have souls.  You might substitute Infidel for Christian, as the answer. 

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The answer would be B. the soul passes through divine spheres.

Sufis believe that the soul passes through divine spheres.

Hope this Helps ;P

There are: 126   156   186   216   246   276   306   336   366   396   426   456   486   516   546   576   606   636   666   696   726   756   786   816   Total =  24

You have: 999 - 100 + 1 =900 - 3-digit numbers

Probability =24 / 900 = 2 / 75

Than swer is D. Sufism.

Tariqa is a religios order of Sufism.

Hope this Helps ;P

Here are the last 20 digits:  9^2004 mod 10^20 =2025682304 6793686561

r u in AOPS???

no wonder she got mad

9^0  01

9^1  09

9^2  81

   729

   6561

   59049

   531441

   4782969

  43046721

  387420489

  3486784401

  31381059609

Lo0ks like if even   the number ends in 1

  the tens pattern as 0  8  2  6  4       4  6  2   8  0      2004   would be the 5th in the sequence

 1 + 6 = 7          My guess

You know she dosen't like to be called that? XD

thank Pinky

The answer would be C. Mongols.

An invasion by the mongols in the thirteenth century marked the downfall of the Seljuk dynasty.

Hope this Helps ;P

Thanks a mill

A Jerusalem

V ≈ 1231.5in³

OK....no prob   !!!!

Wow, thanks so much Cphill this was really helpful!

OK, HL.....

The salaries could be as much as $4500 less than $37,500  which means that the salaries could be as little as 

$37,500 -$4500  = $33,000  

Or..

The salaries could be as much as  $4500 more  than $37,500  = $37,500 + $4500  =  $42,000

So

The salries could range from $33,000 to $42000   and  lx - 37,5000 l ≤ $4500   is correct  [ the first answer ]

"In this problem, a and b are positive integers.
When a is written in base 9, its last digit is 5.
When b is written in base 6, its last two digits are 53.
When a-b is written in base 3, what are its last two digits? Assume a-b is positive."

In base 10 the last two digits of a make:         a1*9 + 5

In base 10 the last two digits of b make:        5*6 + 3 = 33

In base 10 the last two digits of a-b make:    d1*3 + d0

So:  a1*9 + 5 - 33 = d1*3 + d0  or a1*9 - 28 = d1*3 + d0  where 0<= a1 <= 8 and 0<= d1,d0 <= 2

Hence possible values for a1*9 - 28 are:  -28    -19    -10    -1    8    17    26    35    44 (just letting a1 run from 0 to 8)

The maximum value of d1*3 + d0 is 8, the minimum value is 0, because 0<= d1,d0 <= 2, and a-b can’t be negative, so the only option is d1*3 + d0 = 8

which means that d1 = 2 and d0 = 2. 

i.e. the last two digits of a - b in base 3 are 22

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For a specific answer, you need to present one or more equations in the context of your question.

Generally, for continuous, linear function equations the domain and range will be all real numbers, except when the denominator becomes zero (0).

Let A be a matrix, and let x and y be linearly independent vectors such that
\(Ax=y,\ Ay=x+2y\)
Then we have that
\(A^5x=ax+by\)  for some scalars of a and b. Find the ordered pair (a,b).

\(\begin{array}{|rcll|} \hline \mathbf{Ax} &=&y \quad & | \quad \cdot A \\ A^2x &=& Ay \quad & | \quad \mathbf{Ay =x+2y} \\ &=& x+2y \quad & | \quad \cdot A \\\\ A^3x &=& Ax+2 Ay \\ &=& y+2(x+2y) \\ &=& 2x+5y \quad & | \quad \cdot A \\\\ A^4x &=& 2Ax+5 Ay \\ &=& 2y+5(x+2y) \\ &=& 5x+12y \quad & | \quad \cdot A \\\\ A^5x &=& 5Ax+12 Ay \\ &=& 5y+12(x+2y) \\ &=& \mathbf{12x+29y} \\ \hline \end{array}\)

By intercept theorem,

\(\dfrac{\text{AC}}{\text{DA} }= \dfrac{\text{BC}}{\text{BE}}\\ \dfrac{15}{15-10} = \dfrac{8 + \text{BE}}{\text{BE}}\\ 15\text{BE} = 40 + 5\text{BE}\\ 10\text{BE} = 40\\ \text{BE}=4\)

\(\begin{cases} a + ab^2 = 40b \;\;\;\!\;\!--- (1)\\ a-ab^2 = -32b --- (2)\\ \end{cases}\\ (1) + (2) : 2a = 8b\\ b = \dfrac{a}{4}\\ \text{Substitute }b = \dfrac{a}4\text{ into (1),}\\ a + a \cdot \left(\dfrac{a^2}{16}\right) = 40\left(\dfrac{a}{4}\right)\\ a^3 -144a = 0\\ a(a+12)(a-12) = 0\\ a \in\{ 0, 12, -12\}\)

3)

\(\text{When }f_1 = x,\\ f_2 = f_1(f_1) = f(x) = f_1(x) = x\\ f_3 = f_2(f_1) = f(x) = x\\ \text{When } f_k = x, \\f_{k+1} = f_k(f_1) = f(x) = f_1 (x) = x\\ \implies \text{When }f_1(x) = x, f_n (x) = x \quad \forall x \in \mathbb Z^+\\ \implies \text{The set }S \text{ only contains the solutions to the equation } f_1(x) = x.\\ \text{When }f_1(x) = x,\\ \dfrac{x+6}{x} = x\\ x^2 - x - 6 = 0\\ (x-3)(x + 2) = 0\\ x = 3 \vee x = -2\\ card(S) = 2\)