answer:Let's calculate how much money Stormi has made so far by washing cars. She washed 3 cars for 10 each, so she made: 3 cars * 10/car = 30 from washing cars. She needs 24 more to afford the bicycle that costs 80, so she has already saved: 80 - 24 = 56. Since she made 30 from washing cars, the rest must have come from mowing lawns. So, she made: 56 - 30 = 26 from mowing lawns. Now, let's find out how many lawns she mowed. She earns 13 for each lawn, so the number of lawns she mowed is: 26 / 13/lawn = 2 lawns. Therefore, Stormi mowed boxed{2} lawns.

answer:1. Given the natural number (x = 6^n + 1), where (n) is an odd natural number. 2. It is known that (x) has exactly three distinct prime divisors and one of those prime divisors is 11. To solve for (x): 3. Since (x) is divisible by 11, (n) must be an odd multiple of 5 (as (6^5 equiv 1 pmod{11})). Therefore, we can write (x) as (6^{5m} + 1) for some odd (m). 4. Using the formula for the sum of a geometric series, we have: [ x = 6^{5m} + 1 = (6^m)^5 + 1 ] Expressing (x) in terms of products of (p) and (q): [ x = (6^m + 1) cdot (6^{4m} - 6^{3m} + 6^{2m} - 6^m + 1) = p cdot q ] where (p = 6^m + 1) and (q = 6^{4m} - 6^{3m} + 6^{2m} - 6^m + 1). 5. Note that (p) and (q) are coprime. Suppose a number (r) divides both (p) and (q). Then: [ r mid 6^m + 1 ] [ r mid 6^{4m} - 6^{3m} + 6^{2m} - 6^m + 1 ] Given (p = 6^m + 1): [ q - p rightarrow 6^{4m} - 6^{3m} + 6^{2m} - 6^m quad text{(also divisible by (r))}] However, since (6^m equiv -1 pmod{r}): [ 6^{2m} equiv 1 pmod{r} ] [ 6^{4m} equiv 1 pmod{r} ] Thus: [ r mid (6^{4m} - 6^{3m} + 6^{2m} - 6^m + 1) = 5 ] Hence, (r) must be 1 or 5. We check modulo 5: Since (p = 6^m + 1 equiv 2^m + 1 neq 0 pmod{5}) for any odd (m): So, (r) must be 1. Finally: 6. Since (m) is odd, and one of the prime divisors of (p) should be 7: - If (m = 1), (p = 6^1 + 1 = 7). - If higher odd values, (6^m + 1) increases and is not a power of 7. Additionally, (p) must not be divisible by powers of 7 by checking divisibility. 7. Evaluate: If (m = 1): [ x = 6^5 + 1 = 7776 + 1 = 7777 ] And: [ 7777 = 7 cdot 11 cdot 101 ] There are exactly three distinct prime factors. Conclusion, the solution satisfies the given condition. [ boxed{7777} ]

answer:Since alpha is an angle in the fourth quadrant, we have k cdot 360^circ + 270^circ < alpha < k cdot 360^circ + 360^circ, where k in mathbb{Z}. Therefore, -k cdot 360^circ - 360^circ < -alpha < -k cdot 360^circ - 270^circ, where k in mathbb{Z}. Let n = -k, where n in mathbb{Z}, thus, we have n cdot 360^circ - 360^circ < -alpha < n cdot 360^circ - 270^circ, where n in mathbb{Z}. Hence, the terminal side of -alpha is in the first quadrant. Therefore, the correct answer is: boxed{text{A}}. **Analysis:** This problem involves representing the fourth quadrant angle alpha with inequalities, then using the properties of inequalities to determine the inequalities that -alpha satisfies, thereby determining the quadrant in which the terminal side of angle -alpha lies.

answer:Let the x-coordinates of the vertices of P_1 be x_1, x_2, ..., x_{150}. By the midpoint formula, the x-coordinates of the vertices of P_2 are frac{x_1+x_2}{2}, frac{x_2+x_3}{2}, ..., frac{x_{150}+x_1}{2}. The sum of these x-coordinates equals: frac{2(x_1 + x_2 + ... + x_{150})}{2} = x_1 + x_2 + ... + x_{150}. Since the sum of the x-coordinates of P_1 is 3010, then the sum of the x-coordinates of P_2 also equals 3010. Similarly, the sum of the x-coordinates of the vertices of P_3, which are found by applying the midpoint formula again to P_2, will equal the sum of the x-coordinates of P_2. Therefore, the sum of the x-coordinates of P_3 also equals 3010. Conclusion: boxed{3010}