answer:1. We begin by noting that the microorganism moves at a speed of frac{1}{6} mm/sek. Over a period of 1 minute (or 60 seconds), the total distance that a microorganism travels will be: [ text{Distance} = 60 times frac{1}{6} = 10 text{ mm} ] 2. The triangle with sides 30 mm, 40 mm, and 50 mm is a right-angled triangle (since (30^2 + 40^2 = 50^2)). 3. The radius of the inscribed circle in a right-angled triangle can be calculated using the formula: [ r = frac{a + b - c}{2} ] where (a = 30) mm, (b = 40) mm, and (c = 50) mm. Plugging in these values: [ r = frac{30 + 40 - 50}{2} = frac{20}{2} = 10 text{ mm} ] Thus, the radius of the inscribed circle is 10 mm. 4. This implies that all points within the triangle are within 10 mm of its sides. This means that the microorganisms will cover the entire interior of the triangle during their chaotic movement. 5. When moving outward, the microorganisms can reach areas beyond the sides of the triangle up to 10 mm away. This adds an extra region around the triangle. 6. Specifically, the area occupied includes: - The area of the triangle itself, - Three strips (one along each side) each 10 mm wide, plus - The sectors at the vertices, which combine to form a complete circle of radius 10 mm. 7. Calculating the areas, we have: - **Area of the triangle**: [ text{Area}_{text{triangle}} = frac{1}{2} times 30 times 40 = 600 text{ mm}^2 ] - **Area of the strips**: The combined length of the sides of the triangle (i.e., the perimeter (P)) is: [ P = 30 + 40 + 50 = 120 text{ mm} ] The total area covered by the strips is: [ text{Area}_{text{strips}} = 10 times 120 = 1200 text{ mm}^2 ] - **Area of the circular sectors**: These sectors combine to form a circle of radius 10 mm, so: [ text{Area}_{text{sectors}} = pi times (10^2) = 100pi text{ mm}^2 ] 8. Adding the areas together: [ text{Total Area} = 600 + 1200 + 100pi ] 9. Approximating (pi approx 3.14): [ text{Total Area} approx 600 + 1200 + 100 times 3.14 = 600 + 1200 + 314 = 2114 text{ mm}^2 ] # Conclusion The area of the colored surface after 1 minute is approximately (2114) mm(^2): [ boxed{2114 text{mm}^2} ]

answer:Given that the line segment MN is parallel to the y-axis and has a length of 4, and the coordinates of point M are (-1,3), we aim to find the coordinates of point N. 1. Since MN is parallel to the y-axis, the abscissa (x-coordinate) of point N must be the same as that of point M. Therefore, the x-coordinate of N is -1. 2. Given that MN = 4, and considering the direction of the y-axis, point N can be either above or below point M on the coordinate plane. 3. If point N is above point M, then the ordinate (y-coordinate) of N is 3 + 4 = 7. Thus, the coordinates of N would be (-1, 7). 4. If point N is below point M, then the ordinate of N is 3 - 4 = -1. Thus, the coordinates of N would be (-1, -1). Therefore, the coordinates of point N are boxed{(-1,-1) text{ or } (-1,7)}. 【Analysis】 By understanding the properties of lines parallel to the y-axis, we deduce that the abscissa of point N remains unchanged from point M. The length of segment MN provides the difference in their ordinates. Since the question does not specify whether N is above or below M, both possibilities must be considered. 【Comments】 This problem tests the understanding of coordinate geometry, specifically the implications of parallelism with the coordinate axes and distance between points. It's crucial to remember that without specific direction, points parallel to an axis can exist in two locations, leading to two possible answers.

answer:If Hannah's AdBlock blocks all but 20% of ads, it means that 80% of ads are blocked. Therefore, only 20% of ads are not blocked. Out of the 20% of ads that are not blocked, 20% of them are actually interesting. To find out the percentage of ads that are not interesting, we subtract the interesting ads from the total unblocked ads: 100% (of unblocked ads) - 20% (interesting unblocked ads) = 80% (uninteresting unblocked ads) Now, we need to calculate 80% of the 20% of ads that are not blocked to find the percentage of ads that are not interesting and don't get blocked: 80% of 20% = 0.80 * 20% = 16% So, boxed{16%} of the total ads are not interesting and don't get blocked by Hannah's AdBlock.

answer:We begin by proving a lemma involving the relationship between real numbers and products. 1. **Lemma:** Let (a_1, a_2, ldots, a_n, b_1, b_2, ldots, b_n in mathbb{R}_+) (where (mathbb{R}_+) denotes the set of positive real numbers) and let (n geq 2). If (a_i > b_i) for all (1 leq i leq n), then: [ prod_{i=1}^{n} left(a_i - b_iright) leq left( sqrt[n]{prod_{i=1}^{n} a_i} - sqrt[n]{prod_{i=1}^{n} b_i} right)^n. ] 2. **Proof for (n = 2^k) (Induction on (k)):** - **Base Case: (k = 1, n = 2):** [ left(a_1 - b_1right)left(a_2 - b_2right) leq left(sqrt{a_1 a_2} - sqrt{b_1 b_2}right)^2. ] By expanding both sides, this inequality becomes: [ a_1 b_2 + a_2 b_1 - 2 sqrt{a_1 a_2 b_1 b_2} geq 0. ] This reduces to: [ left( sqrt{a_1 b_2} - sqrt{a_2 b_1} right)^2 geq 0, ] which is always true. - **Inductive Step: Assume true for (k = s). Prove for (k = s + 1, n = 2^{s+1}):** Assuming the lemma holds for (n = 2^s): [ prod_{i=1}^{2^s} left(a_i - b_iright) leq left( sqrt[2^s]{prod_{i=1}^{2^s} a_i} - sqrt[2^s]{prod_{i=1}^{2^s} b_i} right)^{2^s}. ] We need to show it holds for (n = 2 cdot 2^s): [ prod_{i=1}^{2^{s+1}} left(a_i - b_iright) = left[ prod_{i=1}^{2^s} left(a_i - b_iright) right] left[ prod_{i=1}^{2^s} left(a_{i+2^s} - b_{i+2^s}right) right]. ] Applying the induction hypothesis: [ leq left( sqrt[2^s]{prod_{i=1}^{2^s} a_i} - sqrt[2^s]{prod_{i=1}^{2^s} b_i} right)^{2^s} left( sqrt[2^s]{prod_{i=1}^{2^s} a_{i+2^s}} - sqrt[2^s]{prod_{i=1}^{2^s} b_{i+2^s}} right)^{2^s}. ] This simplifies to: [ leq left( sqrt[2^{s+1}]{prod_{i=1}^{2^{s+1}} a_i} - sqrt[2^{s+1}]{prod_{i=1}^{2^{s+1}} b_i} right)^{2^{s+1}}. ] Therefore, the lemma holds for all (n = 2^k). 3. **Proof for General (n):** Given that the lemma holds for (n = r + 1), it also holds for (n = r (r geq 2, r in mathbb{Z}_+)). If (A = sqrt[r]{prod_{i=1}^{r} a_i}) and (B = sqrt[r]{prod_{i=1}^{r} b_i}), then: [ prod_{i=1}^{r+1} left(a_i^* - b_i^*right) leq (A - B)^{r+1}. ] Dividing by (A - B): [ prod_{i=1}^{r} left(a_i - b_iright) leq (A - B)^r. ] 4. **Application to Problem:** We are given (9a_i > 11a_{i+1}^2 implies a_i > a_{i+1}^2). 5. **Considering (a_{2016} - a_1^2 leq 0):** If (a_{2016} - a_1^2 leq 0), then the product evaluates to zero: [ M = prod_{i=1}^{2016} left(a_i - a_{i+1}^2right) leq 0. ] 6. **Considering (a_{2016} - a_1^2 > 0):** [ prod_{i=1}^{2016} (a_i - a_{i+1}^2) > 0 implies sqrt{ prod_{i=1}^{2016} a_i } < 1. ] Using the previously proved lemma: [ M leq left(S - S^2 right)^{2016} = (S(1-S))^{2016} leq left(frac{1}{2}right)^{2016} leq frac{1}{4^{2016}}. ] 7. **Choosing ( a_1 = a_2 = cdots = a_{2016} = frac{1}{2} ):** [ M = frac{1}{4^{2016}}. ] Overall, the maximum value that satisfies the given constraints is: [ boxed{frac{1}{4^{2016}}} ]