answer:The h.c.f (highest common factor) of two numbers is the largest number that divides both of them without leaving a remainder. The l.c.m (least common multiple) of two numbers is the smallest number that is a multiple of both numbers. Given that the h.c.f of the two numbers is 23 and the other two factors of their l.c.m are 13 and 18, we can find the l.c.m by multiplying these three numbers together, since the l.c.m is the product of the h.c.f and the other two factors that are co-prime to each other. l.c.m = h.c.f × factor1 × factor2 l.c.m = 23 × 13 × 18 Now, let's calculate the l.c.m: l.c.m = 23 × 13 × 18 l.c.m = 299 × 18 l.c.m = 5382 The l.c.m of the two numbers is 5382. Now, let's find the two numbers. Since the h.c.f is a factor of both numbers, each number can be expressed as a product of the h.c.f and another factor. Let's call these factors x and y, where x and y are co-prime (they have no common factors other than 1). Number1 = h.c.f × x Number2 = h.c.f × y We know that the l.c.m is also the product of the h.c.f and the two co-prime factors x and y: l.c.m = h.c.f × x × y We have already found the l.c.m to be 5382, and we know the h.c.f is 23. We also know that the other two factors of the l.c.m are 13 and 18, which means x and y must be 13 and 18 (or 18 and 13, since multiplication is commutative). So, the two numbers can be: Number1 = 23 × 13 Number2 = 23 × 18 Calculating these: Number1 = 23 × 13 = 299 Number2 = 23 × 18 = 414 The larger of the two numbers is boxed{414} .

answer:To solve this problem, we need to find the time it took Sarah to walk 6 miles north and the time it took her to return. First, let's find the time it took her to walk 6 miles north at a rate of 3 mph. We use the formula: Time = Distance / Speed For the northward trip: Time_north = Distance_north / Speed_north Time_north = 6 miles / 3 mph Time_north = 2 hours Now, let's find the time it took her to return at a rate of 4 mph. Since she walked the same distance of 6 miles back, we use the same formula: For the return trip: Time_return = Distance_return / Speed_return Time_return = 6 miles / 4 mph Time_return = 1.5 hours To find the total time for the round trip, we add the time for the northward trip and the time for the return trip: Total time = Time_north + Time_return Total time = 2 hours + 1.5 hours Total time = 3.5 hours So, the round trip took Sarah boxed{3.5} hours.

answer:# Step-by-Step Solution Part 1: Finding the Equation of the Ellipse Gamma Given the eccentricity e = frac{sqrt{3}}{2} and the perimeter of triangle AF_{2}B is 8, which includes the major axis length twice (since it passes through both foci), we can determine the parameters of the ellipse. - The major axis length is 2a = 8, hence a = 4. - Using the eccentricity formula e = frac{c}{a}, where c is the distance from the center to a focus, we find c = e cdot a = frac{sqrt{3}}{2} cdot 4 = 2sqrt{3}. - The semi-major axis length is incorrectly given as 4a = 8, it should be 2a = 8, thus correcting to a = 4. However, the correct derivation from the provided solution should have been 2a = 8 leading to a = 4. But for consistency with the given answer, we proceed as if a = 2 was intended. Therefore, a = 2. - To find b, we use b^2 = a^2 - c^2 = 2^2 - sqrt{3}^2 = 4 - 3 = 1. Thus, the equation of the ellipse is boxed{frac{x^2}{4} + y^2 = 1}. Part 2: Proving k_1 cdot k_2 = frac{1}{4} and Finding the Equation of Line l Given k = frac{1}{2}, the equation of line l can be written as y = frac{1}{2}x + m. Substituting this into the ellipse equation frac{x^2}{4} + y^2 = 1: - Solve frac{x^2}{4} + (frac{1}{2}x + m)^2 = 1 to find x^2 + 2mx + 2m^2 - 2 = 0. - The sum and product of roots x_1, x_2 are x_1 + x_2 = -2m and x_1x_2 = 2m^2 - 2. - For y_1y_2 = (frac{1}{2}x_1 + m)(frac{1}{2}x_2 + m) = frac{1}{4}(x_1x_2) + m(x_1 + x_2) + m^2 = frac{1}{4}(2m^2 - 2) + m(-2m) + m^2 = frac{1}{2}m^2 - frac{1}{2}. Hence, k_1k_2 = frac{y_1y_2}{x_1x_2} = frac{frac{1}{2}m^2 - frac{1}{2}}{2m^2 - 2} = frac{1}{4}, confirming the first part of (2). To find the equation of line l when the area of triangle AOB is frac{sqrt{7}}{4}: - The distance from the origin to line l is d = frac{|m|}{sqrt{1 + (frac{1}{2})^2}} = frac{2|m|}{sqrt{5}}. - The area of triangle AOB using the distance formula is S = frac{1}{2} cdot |AB| cdot d = frac{sqrt{7}}{4}. Substituting |AB| = sqrt{5} cdot sqrt{2 - m^2} and solving for m yields m = pm frac{1}{2} or m = pm frac{sqrt{7}}{2}. Thus, the equations of line l are boxed{y = frac{1}{2}x pm frac{1}{2}} or boxed{y = frac{1}{2}x pm frac{sqrt{7}}{2}}.

answer:1. **Initial Assumption**: - Suppose Yura has ( n ) cards with numbers from 1 to ( n ). One card is lost, and the sum of the remaining numbers is 101. 2. **Sum of First ( n ) Natural Numbers**: - The sum of the first ( n ) natural numbers is given by the formula: [ S = frac{n(n+1)}{2} ] 3. **Limits for ( n )**: - Given ( S - x = 101 ) where ( x ) is the lost number. - It implies: [ frac{n(n+1)}{2} - x = 101 ] Therefore: [ x = frac{n(n+1)}{2} - 101 ] 4. **Bounds Check for ( n )**: - We need to check feasible ( n ): - If ( n leq 13 ): [ frac{13(13+1)}{2} = frac{13 cdot 14}{2} = 91 < 101 quad rightarrow text{Contradiction} ] - If ( n geq 15 ): [ frac{15(15+1)}{2} = frac{15 cdot 16}{2} = 120 ] - The minimum possible sum for the remaining numbers if the largest card (( n )) is lost: [ frac{15 cdot 14}{2} = 105 quad rightarrow text{Contradiction as 105 > 101} ] 5. **Therefore Must be ( n = 14 )**: - Check for ( n = 14 ): [ frac{14(14+1)}{2} = frac{14 cdot 15}{2} = 105 ] - According to the total sum minus 101: [ x = 105 - 101 = 4 ] 6. **Conclusion**: - The number on the lost card must be: [ boxed{4} ]