answer:The expression 3^{3^{3^{3}}} can be parenthesized in the following ways: 1. 3^{(3^{(3^3)})} 2. 3^{((3^3)^3)} 3. ((3^3)^3)^3 4. (3^{(3^3)})^3 5. (3^3)^{3^3} We will evaluate each case: 1. **3^{(3^{(3^3)})}**: [ 3^{(3^{27})} = 3^{7625597484987} ] This is the original expression given in the problem. 2. **3^{((3^3)^3)}**: [ 3^{(27^3)} = 3^{19683} ] 3. **((3^3)^3)^3**: [ (27^3)^3 = 19683^3 ] 4. **(3^{(3^3)})^3**: [ (3^{27})^3 = 7625597484987^3 ] This is different from case 3. 5. **(3^3)^{3^3}**: [ 27^{3^3} = 27^{27} ] From the evaluations, we see that there are three distinct values: 3^{7625597484987}, 3^{19683}, and 19683^3. Therefore, there are two other possible values aside from the original one. Thus, the answer is textbf{(C) 2}. The final answer is boxed{textbf{(C) } 2}

answer:We start by simplifying the expression: [ frac{frac{1}{x+y}}{frac{1}{x-y}} = frac{1}{x+y} cdot frac{x-y}{1} = frac{x-y}{x+y} ] Substitute the values of x and y: - x = 4 - y = 5 So, [ frac{4-5}{4+5} = frac{-1}{9} = boxed{-frac{1}{9}} ] Conclusion: With the given values x = 4 and y = 5, the computation is straightforward and leads to the exact expression of the fraction, yielding a valid and correct result -frac{1}{9}.

answer:First, identify the common factors in the numerator and denominator. The numbers 15, 45, and 75 have a common factor of 15, and the terms b^4 and b^3 have b^2 as a common factor in all terms. Starting with the numbers: - frac{15b^4 - 45b^3}{75b^2} = frac{15(b^4 - 3b^3)}{75b^2} = frac{15 cdot b^2(b^2 - 3b)}{75b^2}. Now cancelling the common factors: - = frac{cancel{15} cdot cancel{b^2}(b^2 - 3b)}{5 cdot cancel{15} cdot cancel{b^2}} = frac{b^2 - 3b}{5}. Replace b with 2: - = frac{(2^2) - 3(2)}{5} = frac{4 - 6}{5} = frac{-2}{5}. Thus, the simplified value is boxed{-frac{2}{5}}.

answer:We are given a triangle ( triangle ABC ) with vertices ( A, B, ) and ( C ). We need to determine the distance ( B'C' ) where ( B' ) and ( C' ) are the points where the perpendiculars from ( A ) to ( AB ) and ( AC ) respectively, intersect ( BC ). Step 1: Use of the Sine Rule in Triangles ( ABC' ) and ( ACB' ) By applying the Sine Rule in triangles ( triangle ABC' ) and ( triangle ACB' ), we get: [ BC' = frac{c cos alpha}{cos gamma} quad text{and} quad CB' = frac{b cos alpha}{cos beta} ] Step 2: Expression for ( B'C' ) Using the results from Step 1, the distance ( B'C' ) can be expressed as: [ B'C' = BC' + a + CB' = a + frac{b cos alpha}{cos beta} + frac{c cos alpha}{cos gamma} ] Step 3: Simplify the Expression for ( B'C' ) We can factor out common terms and simplify: [ B'C' = a + frac{b cos alpha}{cos beta} + frac{c cos alpha}{cos gamma} ] [ = frac{a cos beta cos gamma + b cos alpha cos gamma + c cos alpha cos beta}{cos beta cos gamma} ] [ = frac{a cos beta cos gamma + cos alpha (b cos gamma + c cos beta)}{cos beta cos gamma} ] Since ( a cos alpha = b cos beta cos gamma + c cos beta cos gamma ): [ B'C' = frac{a cos alpha + a cos beta cos gamma}{cos beta cos gamma} ] [ = frac{a (cos alpha + cos beta cos gamma)}{cos beta cos gamma} ] Using the trigonometric identity ( cos(alpha + beta) = cos alpha cos beta - sin alpha sin beta ): [ = a frac{sin beta sin gamma}{cos beta cos gamma} = a tan beta tan gamma ] Conclusion: The distance ( B'C' ) is given by: [ boxed{a tan beta tan gamma} ] Find the Circumradius ( r ) of ( triangle AB'C' ): Let the circumradius be ( r ), and the center of circumcircle be ( O ). Step 1: Area of ( triangle AB'C' ) Using the product of sines for the area ( Delta ) and rearranging, we get: [ frac{1}{2} B'C' cdot r = frac{1}{2} Y_{perp} (C'B') K ] [ B'C'r = frac{B'C' sin B'}{sin left(B' + C'right)} cdot frac{B'C' sin C'}{sin left(B' + C'right)} sin left(B' + C'right) ] [ r = frac{B'C' sin left(45^circ - frac{beta}{2}right) sin left(45^circ - frac{gamma}{2}right)}{cos left(frac{beta + gamma}{2}right)} ] Since: [ B' = 90^circ - beta, quad C' = 90^circ - gamma, ] We substitute these into the expression for ( r ): [ r = frac{a tan beta tan gamma sin left(45^circ - frac{beta}{2}right) sin left(45^circ - frac{gamma}{2}right)}{cos frac{beta + gamma}{2}} ] Conclusion: The circumradius ( r ) is given by: [ boxed{a tan beta tan gamma sin left(45^circ - frac{beta}{2}right) sin left(45^circ - frac{gamma}{2}right)} ]