answer:Given: A right triangle with perimeter ( k = 40 , mathrm{cm} ) and the radius of the inscribed circle ( varrho = 3 , mathrm{cm} ). 1. Determine the semi-perimeter ( s ): [ s = frac{k}{2} = frac{40 , mathrm{cm}}{2} = 20 , mathrm{cm} ] 2. The relationship between the semi-perimeter ( s ), the radius of the inscribed circle ( varrho ), and the lengths divided by the inscribed circle: [ varrho + x + y = s ] Rearranging this: [ x + y = s - varrho = 20 , mathrm{cm} - 3 , mathrm{cm} = 17 , mathrm{cm} ] 3. By the Pythagorean theorem in terms of the sides where ( a = varrho + x ), ( b = varrho + y ), and the hypotenuse ( c = x + y ): [ (a)^2 + (b)^2 = (c)^2 ] Using ( a = varrho + x ) and ( b = varrho + y ) respectively: [ (varrho + x)^2 + (varrho + y)^2 = (x + y)^2 ] 4. Solving for ( x ) and ( y ) yields the quadratic equation: [ xy = varrho(s - varrho) = 3 , mathrm{cm} times 20 , mathrm{cm} = 60 , mathrm{cm^2} ] 5. Combine these relationships into a quadratic equation in terms of ( z ) (where ( z ) represents ( x ) or ( y )): [ z^2 - (s - varrho)z + varrho s = 0 ] Substituting ( s = 20 , mathrm{cm} ) and ( varrho = 3 , mathrm{cm} ): [ z^2 - 17z + 60 = 0 ] 6. Solving the quadratic equation: [ z = frac{(s - varrho) pm sqrt{(s - varrho)^2 - 4 varrho s}}{2} ] [ z = frac{17 pm sqrt{289 - 240}}{2} ] [ z = frac{17 pm sqrt{49}}{2} ] [ z = frac{17 pm 7}{2} ] [ z_1 = frac{24}{2} = 12 , mathrm{cm}, quad z_2 = frac{10}{2} = 5 , mathrm{cm} ] 7. Now, for the sides of the triangle: [ a = varrho + x = 3 , mathrm{cm} + 12 , mathrm{cm} = 15 , mathrm{cm} ] [ b = varrho + y = 3 , mathrm{cm} + 5 , mathrm{cm} = 8 , mathrm{cm} ] [ c = x + y = 17 , mathrm{cm} ] Conclusion: [ boxed{a = 15 , mathrm{cm}, , b = 8 , mathrm{cm}, , c = 17 , mathrm{cm}} ]

answer:The bank account compounds quarterly at an interest rate of 8/4 = 2 percent per quarter. Therefore, in the course of a year, the bank account compounds as follows: [ (1 + 0.02)^4 = 1.02^4 = 1.08243216 dots ] Converting this compounded value back to an equivalent annual rate: [ 1.08243216 - 1 = 0.08243216 ] To find the percentage, multiply by 100: [ 0.08243216 times 100 = 8.24 ] Therefore, rounding to the nearest hundredth, the interest rate is boxed{8.24} percent.

answer:Given the equations: [x^3 = 3y^2x + 5 - sqrt{7},] [y^3 = 3x^2y + 5 + sqrt{7}.] 1. **Rearrange the Equations**: [ x^3 - 3xy^2 = 5 - sqrt{7} quad text{(i)} ] [ y^3 - 3x^2y = 5 + sqrt{7} quad text{(ii)} ] 2. **Square Both Equations**: Square equation (i): [ (x^3 - 3xy^2)^2 = (5 - sqrt{7})^2 ] Expanding the left hand side: [ (x^3 - 3xy^2)^2 = x^6 - 6x^4y^2 + 9x^2y^4 ] Expanding the right hand side: [ (5 - sqrt{7})^2 = 25 - 2 cdot 5 cdot sqrt{7} + 7 = 32 - 10 sqrt{7} ] Therefore: [ x^6 - 6x^4y^2 + 9x^2y^4 = 32 - 10 sqrt{7} quad text{(iii)} ] Square equation (ii): [ (y^3 - 3x^2y)^2 = (5 + sqrt{7})^2 ] Expanding the left hand side: [ (y^3 - 3x^2y)^2 = y^6 - 6x^2y^4 + 9x^4y^2 ] Expanding the right hand side: [ (5 + sqrt{7})^2 = 25 + 2 cdot 5 cdot sqrt{7} + 7 = 32 + 10 sqrt{7} ] Therefore: [ y^6 - 6x^2y^4 + 9x^4y^2 = 32 + 10 sqrt{7} quad text{(iv)} ] 3. **Add Equations (iii) and (iv)**: [ x^6 - 6x^4y^2 + 9x^2y^4 + y^6 - 6x^2y^4 + 9x^4y^2 = 32 - 10 sqrt{7} + 32 + 10 sqrt{7} ] Simplify: [ x^6 + y^6 + 3x^4y^2 + 3x^2y^4 = 64 ] The left-hand side can be recognized as: [ (x^2 + y^2)^3 = 64 ] 4. **Solve for (x^2 + y^2)**: [ (x^2 + y^2)^3 = 64 ] Taking the cube root of both sides yields: [ x^2 + y^2 = sqrt[3]{64} = 4 ] # Conclusion: [ boxed{4} ]

answer:We start by defining: [ N = left[left(n^2 + nright)^2 - left(n^2 - nright)^2right](n^6 - 1) ] First, notice that: [ left(n^2 + nright)^2 - left(n^2 - nright)^2 ] is a difference of squares, which can be factored as: [ left[left(n^2 + nright)^2 - left(n^2 - nright)^2right] = left[(n^2 + n) - (n^2 - n)right]left[(n^2 + n) + (n^2 - n)right] ] Simplifying inside the factors, we get: [ = left[2nright]left[2n^2right] = 4n(n^2+n^2) = 4n cdot 2n^2 = 8n^3 ] Therefore, we can rewrite N as follows: [ N = 8n^3 (n^6 - 1) ] Next, let’s simplify the second factor: [ N = 8n^3 (n^3 - 1)(n^3 + 1) ] Thus: [ N = 8n^3 (n^3 - 1) n^3 (n^3 + 1) = 8n^6 (n^3 - 1) (n^3 + 1) ] We need to check whether N is divisible by 2016: [ 2016 = 2^5 times 3^2 times 7 ] Let's verify the factors: 1. **Divisibility by (7):** Let ( n = 7k + r ) where ( r ) is the remainder when ( n ) is divided by 7. Thus: [ n^3 equiv r^3 pmod{7} ] where ( r = 0, 1, 2, 3, 4, 5, 6 ). The possible values for ( r^3pmod{7} ) are: [ 0^3 equiv 0,quad 1^3 equiv 1,quad 2^3 equiv 8 equiv 1,quad 3^3 equiv 27 equiv 6, quad 4^3 equiv 64 equiv 1, quad 5^3 equiv 125 equiv 6, quad 6^3 equiv 216 equiv 6 ] Thus, ( n^3 pmod{7} ) could be ( 0, 1, 6 ). Therefore: [ n^3 - 1 text{ or } n^3 + 1 text{ or } n^3 text{ is divisible by 7} ] Hence, ( N ) is divisible by ( 7 ). 2. **Divisibility by (3^2 = 9):** Let ( n = 3k + r ) where ( r = 0, 1, 2 ). Let's expand for each ( r ): - If ( r = 0 ): [ n = 3k implies n^3 = (3k)^3 = 27k^3 ] Clearly, ( 9 mid n^3 ). - If ( r = 1 ): [ n = 3k+1 implies n^3 = (3k+1)^3 = 27k^3 + 27k^2 + 9k+1 ] As ( 9 mid (n^3 - 1) ). - If ( r = 2 ): [ n = 3k+2 implies n^3 = (3k+2)^3 = 27k^3 + 54k^2 + 36k + 8 ] Clearly, ( 9 mid (n^3 + 1) ). Hence, for any integer ( n ), ( 9 mid N ). 3. **Divisibility by (2^5 = 32):** Let ( n = 2k ) or ( n = 2k + 1 ), let's examine each case: - If ( n = 2k ): [ n^3 = (2k)^3 = 8k^3 ] Clearly, ( 32 mid 8n^3 ). - If ( n = 2k + 1 ): For the terms ( (n^3 - 1) ) and ( (n^3 + 1) ), observe: [ n^3 - 1 text{ and } n^3 + 1 text{ are two consecutive even numbers, one divisible by } 2, text{the other by } 4] Hence, ( 32 mid 8n^3 ). Combining the results: [ 2016 = 2^5 times 3^2 times 7 mid N ] Thus, we conclude that: [ boxed{2016 mid N} ]