answer:Let's denote the distance between the man's home and office as ( D ) miles. Since he drives to the office at 20 miles/hr, the time taken to reach the office is ( T_1 = frac{D}{20} ) hours. Let's denote his speed on the way back home as ( S ) miles/hr. The time taken to drive back home is ( T_2 = frac{D}{S} ) hours. The average speed for the entire journey is given by the total distance traveled divided by the total time taken. The total distance traveled is ( 2D ) miles (to the office and back home), and the total time is ( T_1 + T_2 ). The average speed is given as 24 miles/hr, so we can write the equation: [ frac{2D}{T_1 + T_2} = 24 ] Substituting ( T_1 ) and ( T_2 ) with their respective expressions in terms of ( D ) and ( S ), we get: [ frac{2D}{frac{D}{20} + frac{D}{S}} = 24 ] Simplifying the equation, we get: [ frac{2D}{frac{D(20 + S)}{20S}} = 24 ] [ 2D cdot frac{20S}{D(20 + S)} = 24 ] [ frac{40S}{20 + S} = 24 ] Now, we solve for ( S ): [ 40S = 24(20 + S) ] [ 40S = 480 + 24S ] [ 40S - 24S = 480 ] [ 16S = 480 ] [ S = frac{480}{16} ] [ S = 30 ] So, the man's speed while driving back home was boxed{30} miles/hr.

answer:Since a=lg 2 is in the interval (0,1), b=2^{0.5}>1, and c=cos frac{3}{4}pi<0, it follows that c<a<b. Therefore, the answer is: c<a<b. This can be determined by using the monotonicity of exponential and logarithmic functions, as well as trigonometric functions. This problem tests the understanding of the monotonicity of exponential and logarithmic functions, as well as trigonometric functions, and examines reasoning and computational skills. It is considered a basic question. Thus, the final answer is boxed{c<a<b}.

answer:Proof: Since a and b are both positive numbers, to prove a^{a}b^{b} geq a^{b}b^{a}, we only need to prove a^{a-b}b^{b-a} geq 1, which is equivalent to proving (frac{a}{b})^{a-b} geq 1. When a geq b, we have a-b geq 0 and frac{a}{b} geq 1, thus (frac{a}{b})^{a-b} geq 1. When a < b, we have a-b < 0 and 0 < frac{a}{b} < 1, thus (frac{a}{b})^{a-b} > 1. Therefore, for any positive numbers a and b, we have (frac{a}{b})^{a-b} geq 1, with equality holding if and only if a = b. Hence, the statement a^{a}b^{b} geq a^{b}b^{a} is true, with equality holding if and only if a = b. The final result is boxed{a^{a}b^{b} geq a^{b}b^{a}}.

answer:To calculate 98 times 102 using the multiplication formula, we follow these steps: 1. Recognize the pattern as a difference of squares: left(100 - 2right) times left(100 + 2right). 2. Apply the formula a^2 - b^2 = (a - b)(a + b): = 100^2 - 2^2 3. Calculate each term: = 10000 - 4 4. Subtract to find the final answer: = 9996 Thus, the calculation yields boxed{9996}.