answer:At 3:00, the hour hand is at the 3 o'clock position and the minute hand is at the 12 o'clock position. The angle between them is one-fourth of a full revolution, which is frac{1}{4}(360^circ) = 90^circ. Every minute, the minute hand moves frac{360^circ}{60} = 6^circ, and the hour hand moves frac{360^circ}{12 times 60} = 0.5^circ. Therefore, the angle between the hands changes at a rate of 6^circ - 0.5^circ = 5.5^circ per minute. In 20 minutes, the change in angle is 5.5^circ times 20 = 110^circ. Starting from 90 degrees, the angle between the hour and minute hands at 3:20 is 90^circ + 110^circ = 200^circ. Since we need the smaller angle, we subtract this from 360 degrees: 360^circ - 200^circ = 160^circ. Thus, the number of degrees in the smaller angle formed by the hour and minute hands of a clock at 3:20 is boxed{160.0} degrees.

answer:Let ( a ) and ( b ) be natural numbers such that ( a ) is divisible by ( b+1 ) and 43 is divisible by ( a+b ). 1. Since 43 is a prime number, the natural number ( a + b ) must either be 1 (which is not acceptable as they are natural numbers) or 43 itself. [ a + b = 43 ] 2. Given ( b leq 42 ) and ( a = 43 - b ): [ a = 43 - b ] 3. From the condition that ( b + 1 ) is a divisor of ( a ): [ b + 1 mid 44 ] Because ( a + b + 1 = 44 ). 4. The positive divisors of 44 are: [ 1, 2, 4, 11, 22, 44 ] 5. Since ( b ) and ( b ) is a natural number, ( b+1 ) must be one among these divisors, excluding 1 and 44. Therefore, we have: [ b + 1 in {2, 4, 11, 22} ] 6. Solving for ( b ): [ text{If } b + 1 = 2: quad b = 1 quad text{and} quad a = 43 - 1 = 42 ] [ text{If } b + 1 = 4: quad b = 3 quad text{and} quad a = 43 - 3 = 40 ] [ text{If } b + 1 = 11: quad b = 10 quad text{and} quad a = 43 - 10 = 33 ] [ text{If } b + 1 = 22: quad b = 21 quad text{and} quad a = 43 - 21 = 22 ] 7. Verifying each pair ((a, b)): [ (42, 1) quad text{because} quad b+1=2 mid 42 quad text{and} quad 43 mid (42 + 1) ] [ (40, 3) quad text{because} quad b+1=4 mid 40 quad text{and} quad 43 mid (40 + 3) ] [ (33, 10) quad text{because} quad b+1=11 mid 33 quad text{and} quad 43 mid (33 + 10) ] [ (22, 21) quad text{because} quad b+1=22 mid 22 quad text{and} quad 43 mid (22 + 21) ] # Conclusion: (a) ( a ) can be ( 22, 33, 40, ) or ( 42 ). (b) ( b ) can be ( 1, 3, 10, ) or ( 21 ). (boxed{})

answer:To find the altitude of the triangle, we first need to find the area of the square, since the area of the triangle is given to be equal to the area of the square. The area of a square is calculated by squaring the length of one of its sides. So, for a square with a side length of 6 cm, the area (A_square) is: A_square = side^2 A_square = 6 cm * 6 cm A_square = 36 cm² Now that we know the area of the square, which is equal to the area of the triangle, we can use the formula for the area of a triangle to find the altitude (h) of the triangle. The formula for the area (A_triangle) of a triangle is: A_triangle = (base * height) / 2 Since we know the area of the triangle is 36 cm² and the base is 6 cm, we can solve for the height (h): 36 cm² = (6 cm * h) / 2 72 cm² = 6 cm * h h = 72 cm² / 6 cm h = 12 cm Therefore, the altitude of the triangle is boxed{12} cm.

answer:1. **Equation Transformation and Geometric Interpretation**: - Given the set ( A ) defined by the equation: [ x^2 + y^2 = 2x + 2y + 23 ] - We complete the square for both ( x ) and ( y ): [ x^2 - 2x + y^2 - 2y = 23 ] [ (x-1)^2 - 1 + (y-1)^2 - 1 = 23 ] [ (x-1)^2 + (y-1)^2 = 25 ] - This is the equation of a circle with center ((1, 1)) and radius ( 5 ). 2. **Interpretation of Set ( B )**: - The set ( B ) is given by: [ |x-1| + |y-1| = 5 ] - This represents a diamond (or square rotated by 45°) centered at ((1, 1)) with a radius (distance from the center to a vertex) of ( 5 ). 3. **Intersection of Sets ( A ) and ( B )**: - The circle and the square will intersect at four points when ( |x-1| ) and ( |y-1| ) reach their respective maximum extents. - The vertices of the square can be determined as ( (-4, 1), (1, 6), (6, 1), (1, -4) ). 4. **Determine Points ( Y_i ) of Intersection**: - Let ( Y_1, Y_2, Y_3, Y_4 ) be the points ((-4, 1), (1, 6), (6, 1), (1, -4)). 5. **Maximum Product Calculation**: - Take ( X ) as a point on the circle where from symmetry, we consider ( X ) on the arc between ( Y_2 ) and ( Y_3 ) for determination. Let the distance from ( X ) to the line ( Y_1 Y_2 ) be ( a ). - Using symmetry and properties of a circle inscribed in a square, we get: [ X Y_1 cdot X Y_2 = 10a ] - Similarly: [ X Y_3 cdot X Y_4 = 10(5sqrt{2} - a) ] - Therefore, the product: [ X Y_1 cdot X Y_2 cdot X Y_3 cdot X Y_4 = 100a(5sqrt{2} - a) ] 6. **Optimization**: - To find the maximum value of ( 100a(5sqrt{2} - a) ), set the derivative equal to zero: [ frac{d}{da}[100a(5sqrt{2} - a)] = 0 ] [ 100(5sqrt{2} - 2a) = 0 ] [ 5sqrt{2} = 2a ] [ a = frac{5sqrt{2}}{2} ] - Substitute ( a = frac{5sqrt{2}}{2} ) into the product: [ 100 left( frac{5sqrt{2}}{2} right) left(5sqrt{2} - frac{5sqrt{2}}{2}right) ] [ 100 left( frac{5sqrt{2}}{2} right) left(frac{5sqrt{2}}{2}right) ] [ 100 left( frac{25 cdot 2}{4} right) ] [ 1250 ] # Conclusion: [ boxed{1250} ]