answer:Let's denote the speed during the first part of the journey as ( v ) kmph. The first part of the journey lasted 3 hours, so the distance covered during the first part is ( 3v ) km. The second part of the journey is at 60 kmph, and since the total journey time is 5 hours, the second part of the journey lasted ( 5 - 3 = 2 ) hours. Therefore, the distance covered during the second part is ( 60 times 2 = 120 ) km. The total distance covered is the sum of the distances covered in the first and second parts of the journey, which is given as 240 km. So we have: [ 3v + 120 = 240 ] Now, we can solve for ( v ): [ 3v = 240 - 120 ] [ 3v = 120 ] [ v = frac{120}{3} ] [ v = 40 ] So, the speed during the first part of the journey was boxed{40} kmph.

answer:Let's calculate the number of shirts Kendra needs for each type of activity per week and then multiply by three to cover the three-week period. 1. School shirts: 5 weekdays x 1 shirt per day = 5 shirts per week 2. After-school club shirts: 3 days a week x 1 shirt per day = 3 shirts per week 3. Saturday shirts: 1 for workout + 1 for art class + 1 for the rest of the day = 3 shirts per Saturday 4. Sunday shirts: 1 for church + 1 for volunteer work + 1 for the rest of the day = 3 shirts per Sunday Now, let's add these up for one week: 5 (school) + 3 (club) + 3 (Saturday) + 3 (Sunday) = 14 shirts per week Since Kendra wants to do laundry only once every three weeks, we multiply the weekly total by three: 14 shirts per week x 3 weeks = 42 shirts Kendra needs boxed{42} shirts to be able to only do laundry once every three weeks.

answer:1. **Identify Geometry and Setup**: There are four equilateral triangles each with a side of 8 cm, with each consecutive one rotated about its center by increasing angles (0°, 45°, 90°, 135°). 2. **Calculate the Height and Area of One Triangle**: The height (h) of one equilateral triangle is given by ( h = frac{sqrt{3}}{2} times 8 = 4sqrt{3} ) cm. The area (A) of one triangle is ( A = frac{sqrt{3}}{4} times 8^2 = 16sqrt{3} ) square cm. 3. **Overlap and Total Area Calculation**: Given the rotations, there will be overlapping regions, but for simplicity, we calculate the non-overlapping area by considering the full area of each triangle. Since the triangles are stacked and rotated, the overlapping area does not easily subtract from the total in a straightforward manner without complex geometric constructions. Therefore, the total area without considering overlaps is ( 4 times 16sqrt{3} = 64sqrt{3} ) square cm. 4. **Estimate Overlap Adjustment**: Overlaps occur mainly at the center and are roughly similar in size due to symmetry. Estimating the area of one major overlap (approximately a smaller equilateral triangle formed by the intersection, side about half of the original), the area of this smaller triangle is roughly ( frac{sqrt{3}}{4} times 4^2 = 4sqrt{3} ) square cm. Multiplying by the number of main overlaps (3), we have ( 3 times 4sqrt{3} = 12sqrt{3} ) square cm to subtract. 5. **Final Area Calculation**: ( text{Total Area} = 64sqrt{3} - 12sqrt{3} = 52sqrt{3} ) square cm. Conclusion with boxed answer: The area of the resulting geometric figure is 52sqrt{3 text{ square cm}}. The final answer is boxed{B}

answer:To find the domain of the function ( y = log_a(x - ka) + log_a(x^2 - a^2) ), we need to ensure that the arguments of both logarithmic functions are positive. 1. **Argument 1: Ensure ( x - ka > 0 )** [ x - ka > 0 ] [ x > ka ] 2. **Argument 2: Ensure ( x^2 - a^2 > 0 )** [ x^2 - a^2 > 0 ] [ (x - a)(x + a) > 0 ] To solve ( (x - a)(x + a) > 0 ), consider the intervals defined by the roots ( x = a ) and ( x = -a ). Check the sign of the product in each interval: - For ( x < -a ), both ( x - a ) and ( x + a ) are negative, so their product is positive. - For ( -a < x < a ), ( x - a ) is negative and ( x + a ) is positive, so their product is negative. - For ( x > a ), both ( x - a ) and ( x + a ) are positive, so their product is positive. Thus, ( x^2 - a^2 > 0 ) when ( x < -a ) or ( x > a ). 3. **Combine the Constraints:** Given the domain is ( { x mid x > a } ), we need to combine this with ( x > ka ). Therefore, we need: [ x > max(a, ka) ] Since our domain is ( x > a ), the constraint ( x > ka ) must hold for ( x > a ). Therefore: [ max(a, ka) = a implies a > ka ] [ frac{a}{k} > ka implies k in [-1, 1] ] Thus, the range of real values for ( k ) is: [ boxed{[-1, 1]} ]