answer:To find the shaded area outside the circle but inside the semicircles, we need to proceed with the following steps: 1. **Determine the properties of the square**: - The side length of the square ( AB = 2 ). - The diagonal of the square ( AC ) can be calculated using the Pythagorean theorem: [ AC = sqrt{AB^2 + BC^2} = sqrt{2^2 + 2^2} = sqrt{4 + 4} = sqrt{8} = 2sqrt{2} ] 2. **Calculate the area of the circle**: - The diagonal of the square is the diameter of the circle inscribed around the square: [ text{Diameter of the circle} = 2sqrt{2} ] - The radius of the circle is: [ text{Radius} = frac{2sqrt{2}}{2} = sqrt{2} ] - The area of the circle is: [ text{Area of the circle} = pi (sqrt{2})^2 = 2pi ] 3. **Calculate the area of the square**: - The area of the square ( ABCD ) is: [ text{Area of the square} = 2 times 2 = 4 ] 4. **Determine the area outside the square but inside the circle**: - The area outside the square but inside the circle is: [ 2pi - 4 ] 5. **Calculate the area of each semicircle**: - Each side of the square is used as the diameter of a semicircle, hence the radius of each semicircle is: [ text{Radius of each semicircle} = frac{2}{2} = 1 ] - The area of each semicircle is: [ text{Area of each semicircle} = frac{1}{2} pi (1^2) = frac{1}{2} pi ] - The total area of the four semicircles is: [ 4 times frac{1}{2} pi = 2 pi ] 6. **Determine the shaded region**: - The shaded region is the area covered by semicircles not overlapped by the circle: [ text{Shaded area} = 2pi - (2pi - 4) = 2pi - 2pi + 4 = 4 ] # Conclusion The area of the shaded region outside the circle and inside the semicircles is [ boxed{4} ]

answer:Lisa has kept 4 children's spoons and 2 decorative spoons, which adds up to: 4 children x 3 spoons each = 12 spoons 12 spoons (children's) + 2 spoons (decorative) = 14 spoons She has a new set of cutlery with 10 large spoons. So, the total number of spoons she has now is: 10 large spoons (new set) + 14 spoons (kept) = 24 spoons Lisa has a total of 39 spoons now. To find out how many teaspoons are in the new set of cutlery, we subtract the 24 spoons we already accounted for from the total: 39 spoons (total) - 24 spoons (accounted for) = 15 teaspoons So, there are boxed{15} teaspoons in the new set of cutlery.

answer:Let's denote the number of cans Lorenzo had as ( c ). Each can contributed one thumbtack to each of the 120 boards, so each can initially had ( 120 + 30 ) thumbtacks because there were 30 left in each can at the end of the day. So, each can had ( 150 ) thumbtacks at the beginning of the day. The total number of thumbtacks from all the cans was 450, so we can set up the equation: ( c times 150 = 450 ) Now, we solve for ( c ): ( c = frac{450}{150} ) ( c = 3 ) Lorenzo had boxed{3} cans of thumbtacks.

answer:For Manu to win, he must first flip two consecutive heads when his turn comes, and none of the previous players (Juan, Carlos, Ana) should have flipped heads. The probability of not flipping heads in one turn is frac{1}{2}. Manu needs to flip two consecutive heads, which has a probability of left(frac{1}{2}right)^2 = frac{1}{4}. - **Probability for Manu to win on his first opportunity** (after Juan, Carlos, Ana, all tails, and then two heads by Manu): [ left(frac{1}{2}right)^3 cdot frac{1}{4} = frac{1}{32}. ] - **Probability for Manu to win on his second opportunity** (another full cycle of tails followed by two heads by Manu): [ left(frac{1}{2}right)^7 cdot frac{1}{4} = frac{1}{512}. ] - **Continuing this pattern**, the probability that Manu wins on his nth turn follows: [ left(frac{1}{2}right)^{4n} cdot frac{1}{4}. ] - **Summing these probabilities** as a geometric series: [ sum_{n=1}^{infty} frac{1}{32} left(frac{1}{16}right)^{n-1} = frac{frac{1}{32}}{1-frac{1}{16}} = frac{frac{1}{32}}{frac{15}{16}} = frac{1}{30}. ] The probability that Manu will win under these new conditions is: [ boxed{frac{1}{30}}. ]