answer:To analyze each option step-by-step: **Option A**: sqrt{3}+sqrt{6}=sqrt{9} - This operation assumes that the sum of two square roots equals the square root of their sum, which is not a valid algebraic property. Thus, this option is incorrect. **Option B**: 3sqrt{5}-sqrt{5}=2 - Simplifying the left side, we have: 3sqrt{5}-sqrt{5} = (3-1)sqrt{5} = 2sqrt{5}. This does not equal 2, thus this option is incorrect. **Option C**: sqrt{24}÷sqrt{6}=4 - Simplifying the division, we have: sqrt{24}÷sqrt{6} = sqrt{frac{24}{6}} = sqrt{4} = 2. This does not equal 4, thus this option is incorrect. **Option D**: sqrt{3}×sqrt{5}=sqrt{15} - Multiplying the square roots, we use the property sqrt{a} times sqrt{b} = sqrt{ab}, so sqrt{3}×sqrt{5} = sqrt{3 times 5} = sqrt{15}. This matches the given operation, thus this option is correct. Therefore, the correct answer is: boxed{D}.

answer:1. The pattern starts with 3 toothpicks in the first stage. 2. From the second stage onward, 3 more toothpicks are added to the previous stage's total. 3. Therefore, the number of toothpicks in the nth stage can be expressed as: ( T_n = 3 + 3(n-1) ). Plugging in ( n = 20 ): [ T_{20} = 3 + 3 times (20 - 1) ] [ T_{20} = 3 + 3 times 19 ] [ T_{20} = 3 + 57 ] [ T_{20} = boxed{60} ]

answer:(Full marks for this question: 12) (I) Since overrightarrow{a}//overrightarrow{b}, sin αcos α-cos α(cos α-2sin α)=0, 3sin αcos α-cos ^{2}α=0 ...(3 points) As cos α≠ 0, 3sin α-cos α=0, Hence, tan α= frac {1}{3} ...(5 points) (II) From |overrightarrow{b}|=|overrightarrow{c}|, we know that sin ^{2}α+(cos α-2sin α)^{2}=5 ...(6 points) 1-2sin 2α+4sin ^{2}α=5, -2sin 2α+4sin ^{2}α=4. sin 2α+cos 2α=-1. sin (2α+ frac {π}{4})=- frac { sqrt {2}}{2} ...(9 points) Given 0 < α < π, we know that frac {π}{4} < 2α+ frac {π}{4} < frac {9π}{4}, Hence, 2α+ frac {π}{4}= frac {5π}{4} or 2α+ frac {π}{4}= frac {7π}{4} ...(11 points) Therefore, boxed{α= frac {π}{2}} or boxed{α= frac {3π}{4}} ...(12 points)

answer:: To show that points K and L are equidistant from the line AT, we need to use symmetry and properties of the angles and triangles in the given geometric configuration. 1. Notice that angle A is 60^circ: [ angle A = 60^circ ] 2. Point A is the intersection point of the tangent lines to the circumcircle of the triangle at vertex A. This implies the tangent line properties and the angles they subtend. 3. Consider the triangles ABL and ACK. These triangles are similar and symmetrical with respect to the bisector of angle A (which is also the median). 4. Because triangle ABL and triangle ACK are regular triangles (as angle A = 60^circ is a standard angle for equilateral triangles and due to symmetry), then triangle ABC and triangle ALK will be symmetric relative to the bisector of angle A. 5. Since triangles ALK and ABC are symmetric, their medians are also symmetrical in these triangles. As medians split the triangles into two equal halves, we find that AK = AL. 6. Since AKL is symmetrical, the altitude from A to KL serves as the median, implying KT = LT where T is the foot of the perpendicular from A to KL: [ T in text{median} quad Rightarrow quad KT = LT ] This shows that points K and L are equidistant from the line AT. [ boxed{text{Equidistant}} ]