answer:1. To find the integer part of ( A ) where [ A = frac{21 times 62 + 22 times 63 + 23 times 64 + 24 times 65 + 25 times 66}{21 times 61 + 22 times 62 + 23 times 63 + 24 times 64 + 25 times 65} times 199, ] we start by simplifying the numerator and denominator. 2. Write the numerator and denominator in terms of a common Let: [ A = frac{21 times 61 + 21 + 22 times 62 + 62 + 23 times 63 + 63 + 24 times 64 + 64 + 25 times 65 + 65}{21 times 61 + 22 times 62 + 23 times 63 + 24 times 64 + 25 times 65} times 199 ] 3. Group the common terms: [ A = left(1 + frac{21 + 62 + 63 + 64 + 65}{21 times 61 + 22 times 62 + 23 times 63 + 24 times 64 + 25 times 65}right) times 199 ] 4. We now evaluate the fraction: Lower Bound: [ A > 199 + frac{21 + 22 + 23 + 24 + 25}{(21 + 22 + 23 + 24 + 25) times 65} times 199 ] [ A > 199 + frac{115}{115 times 65} times 199 ] [ A > 199 + frac{1}{65} times 199 ] [ A > 199 + frac{199}{65} ] [ A > 199 + frac{199}{65} ] [ A > 199 + 3 frac{4}{65} ] [ A > 199 + 3 + frac{4}{65} ] [ A > 202 frac{4}{65}. ] Upper Bound: [ A < 199 + frac{21 + 62 + 63 + 64 + 65}{21 times 61 + 22 times 62 + 23 times 63 + 24 times 64 + 25 times 65} times 199 ] [ A < 199 + frac{21 + 62 + 63 + 64 + 65}{(21 + 22 + 23 + 24 + 25) times 61} times 199 ] [ A < 199 + frac{275}{115 times 61} times 199 ] [ A < 199 + frac{275}{7015} times 199 ] [ A < 199 + frac{199}{7015} times 275 ] [ A < 199 + frac{199 times 275}{7015} ] [ A < 199 + 3 frac{16}{115} ] [ A < 202 frac{16}{115}. ] 5. Combining both, we get: [ 202 < A < 202 frac{16}{61}. ] Conclusion: The integer part of ( A ) is ( boxed{202} ).

answer:First, simplify the expression by combining like terms: [ (16x^7 + 36x^4 - 9) - (4x^7 - 6x^4 - 9) = 16x^7 + 36x^4 - 9 - 4x^7 + 6x^4 + 9 = (16x^7 - 4x^7) + (36x^4 + 6x^4) + (-9 + 9) = 12x^7 + 42x^4. ] Next, factor out common terms: [ 12x^7 + 42x^4 = 6x^4(2x^3 + 7). ] Thus, the completely factored form of the expression is: [ boxed{6x^4(2x^3 + 7)}. ]

answer:Let's calculate the number of chickens affected by each disease and the number that died as a result. Disease A: - Infected chickens: 15% of 800 = 0.15 * 800 = 120 chickens - Death rate among infected: 45% of 120 = 0.45 * 120 = 54 chickens died After Disease A, the number of remaining chickens is 800 - 54 = 746 chickens. Disease B: - Infected chickens: 20% of the remaining healthy chickens (746) = 0.20 * 746 = 149.2 chickens Since we can't have a fraction of a chicken, we'll round this to 149 chickens. - Death rate among infected: 30% of 149 = 0.30 * 149 = 44.7 chickens died Again, since we can't have a fraction of a chicken, we'll round this to 45 chickens. Total chickens that died due to both diseases: 54 (from Disease A) + 45 (from Disease B) = 99 chickens. Carla bought 12.5 times as many chickens as the number that died due to both diseases: 12.5 * 99 = 1237.5 chickens Since we can't have a fraction of a chicken, we'll round this to 1238 chickens. Now, let's add the number of chickens Carla bought to the remaining chickens after the diseases: Remaining chickens after diseases: 746 - 45 = 701 chickens Total chickens after purchase: 701 + 1238 = 1939 chickens Therefore, Carla has boxed{1939} chickens after buying more.

answer:1. **Observations and Symmetry**: - The point ( C ) is reflected across the line ( KM ) to point ( C_1 ). - Thus, ( CC_1 perp KM ). 2. **Equal Angles Implies Isosceles Triangle**: - From the problem, angles ( AKM ) and ( KMB ) are equal. - This implies that triangles ( CKM ) and ( KMB ) are isosceles with ( C_1C = CC_1 ) and the same base angles. - Therefore, the altitude ( CC_1 ) in this configuration acts as the angle bisector of ( angle ACB ). 3. **Angle Bisector Theorem**: - According to the Angle Bisector Theorem, if ( C_1 ) is the point on ( AB ) where the angle bisector of ( angle ACB ) intersects ( AB ), then: [ frac{AC_1}{BC_1} = frac{AC}{CB} ] - Given ( AC = 5 ) cm and ( BC = 6 ) cm, the ratio is: [ frac{AC_1}{BC_1} = frac{5}{6} ] 4. **Finding Lengths Using Ratios**: - Let ( AC_1 = x ) and ( BC_1 = y ). From ( AB = 7 ) cm, we have: [ x + y = 7 ] and [ frac{x}{y} = frac{5}{6} ] - Solving the ratio equation for ( x ) in terms of ( y ): [ x = frac{5}{6}y ] - Substituting ( x = frac{5}{6}y ) into ( x + y = 7 ): [ frac{5}{6}y + y = 7 ] - Simplify and solve for ( y ): [ frac{5y + 6y}{6} = 7 implies frac{11y}{6} = 7 implies y = frac{42}{11} text{ cm} ] 5. **Calculate ( AC_1 ) and ( BC_1 )**: - Using the value of ( y ), find ( x ): [ x = frac{5}{6} times frac{42}{11} = frac{35}{11} text{ cm} ] 6. **Conclusion**: - Thus, the lengths are: [ AC_1 = frac{42}{11} text{ cm} ] [ BC_1 = frac{35}{11} text{ cm} ] [ boxed{frac{42}{11} text{ cm} text{ and } frac{35}{11} text{ cm}} ]