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A premier international ice hockey tournament featuring top teams from around the world competing against each other in thrilling matches.

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<|diff_marker|> ADD A1000 <|diff_marker|> ADD A1100 <|diff_marker|> ADD A1200 <|diff_marker|> ADD A1300 <|diff_marker|> ADD A1400 <|diff_marker|> ADD A1500 <|diff_marker|1. The area ( A ) (in square feet) of a rectangular parking lot is given by ( A = lw ) where ( l ) is the length (in feet) of one side of the lot and ( w ) is its width (in feet). Which equation represents ( l ) in terms of ( w ) when ( A = 15,000 )? (A) ( l = frac{15,000}{w} ) (B) ( l = frac{w}{15,000} ) (C) ( l = w + frac{15,000}{w} ) (D) ( l = w - frac{15,000}{w} ) (E) ( l = w + frac{15,000}{w} + w ) 2. The amount ( A ), in grams, of a certain radioactive substance present after ( t ) years is given by ( A = 100(0.5)^{t/20} ). Determine the half-life of this substance. (A) twenty years (B) twenty-five years (C) one hundred years (D) forty years (E) four hundred years 3. If ( f(x) = (-x)^3 ), what is ( f(5) )? (A) -125 (B) -25 (C) 25 (D) 125 (E) None of these 4. If ( |a - b| = |b - a| ), which statement about ( a ) and ( b ) must be true? (A) Both ( a ) and ( b ) are positive. (B) The sum of ( a ) and ( b ) is zero. (C) The difference of ( a ) and ( b ) is zero. (D) Either ( a ) or ( b ) is zero. (E) None of these === 1. Given the area formula ( A = lw ), we need to solve for ( l ): [ l = frac{A}{w} ] Substituting ( A = 15,000 ): [ l = frac{15,000}{w} ] So the correct answer is: (A) ( l = frac{15,000}{w} ) 2. The formula for radioactive decay is given by: [ A = A_0left(frac{1}{2}right)^{t/T} ] where ( T ) is the half-life. In this case: [ A = 100left(0.5right)^{t/20} ] Comparing this with the standard form: [ T = 20 text{ years} ] So the correct answer is: (A) twenty years 3. Given ( f(x) = (-x)^3 ), we need to find ( f(5) ): [ f(5) = (-5)^3 = -125 ] So the correct answer is: (A) -125 4. Given ( |a - b| = |b - a| ): Since absolute value properties tell us that: [ |a - b| = |b - a| ] This equation holds true for all real numbers ( a ) and ( b ). Therefore: (E) None of these## Problem: How did economic factors contribute to social unrest during late Roman Britain? ## Explanation: Economic factors played a significant role in contributing to social unrest during late Roman Britain due to several interrelated issues: 1. **Heavy Taxation**: The Roman Empire imposed heavy taxes on its provinces to fund its military campaigns and administrative costs. In late Roman Britain, these taxes placed significant financial burdens on local populations, including small farmers who struggled to meet their obligations. This often led to land confiscations when taxes went unpaid. 2. **Land Ownership Issues**: Wealthy landowners often accumulated large estates by buying up land from poorer farmers who could not afford their taxes or debts. This concentration of land ownership led to social inequality and resentment among dispossessed smallholders who were forced into tenant farming or laboring on large estates under harsh conditions. 3. **Decline in Trade**: As external threats increased during this period, trade routes became less secure due to barbarian invasions along borders like Hadrian’s Wall. This decline disrupted local economies reliant on trade goods such as grain from continental Europe or luxury items from Rome. 4. **Currency Devaluation**: The Roman Empire experienced economic instability partly due to debasement of currency—reducing precious metal content while maintaining face value—which led to inflation. This devaluation reduced purchasing power among ordinary citizens while exacerbating economic inequality. 5. **Military Expenditures**: The need for constant military presence in Britain required substantial resources diverted towards maintaining legions stationed there against potential invasions or internal revolts. This drained local economies further as more wealth was funneled into sustaining military infrastructure rather than addressing local needs. 6. **Disruption from Invasions**: Frequent raids by Picts from Scotland’s northern frontier disrupted agricultural production by destroying crops or forcing communities into fortified settlements where farming was less efficient. 7. **Labor Shortages**: Wars caused demographic changes leading to labor shortages as men were conscripted into military service or died in conflict-related activities; women had fewer partners available leading potentially fewer children born which could cause longer-term demographic impacts on agricultural productivity. 8. **Urban Decay**: As economic pressures mounted alongside external threats like invasions or internal conflicts such as civil wars within Rome itself led urban centers like Londinium experiencing decline; this affected trade networks further destabilizing regional economies dependent upon these urban hubs. In summary: The combination## query ## The arithmetic sequence $(a_i)$ is defined by $a_1 = x$ where $x$ is an unknown positive integer less than $10$, $a_i = a_{i-1} + y$ where $y$ is an unknown common difference that could be either positive or negative but not zero. Given that $a_{11}$ equals $20$, find: 1. The values of $x$. 2. The value(s) of $y$. 3. Finally calculate $a_{20}$ using one valid pair $(x,y)$. ## response ## To solve this problem step-by-step: ### Step 1: Find Possible Values for $x$ We know: [a_{11} = x + 10y] Given: [a_{11} = 20] Thus: [x + 10y = 20] Since $x$ must be an integer between $1$ and $9$, we need values such that: [x + 10y = 20] We can iterate over possible values for $x$: For each value: - If $x=1$: [1 + 10y = 20] [10y = 19] [y = 1.9] which isn't an integer. - If $x=2$: [2 + 10y = 20] [10y =18] [y=1.8] which isn't an integer. - If $x=3$: [3 +10 y=20] [10 y=17] [y=1.7] which isn't an integer. - If $x=4$: [4 +10 y=20] [10 y=16] [y=1.6] which isn't an integer. - If $x=5$: [5 +10 y=20] [10 y=15] [y=1.5] which isn't an integer. - If $x=6$: [6 +10 y=20] [10 y=14] [y=1.4] which isn't an integer. - If $x=7$: [7 +10 y=20] [10 y=13] [y=1.3] which isn't an integer. - If $x=8$: [8 +10 y=20] [10 y=12] [y=1.2] which isn't an integer. - If $x=9$: [9+10 y=20] [10 y=11] [y=1.overline{1}] which isn't an integer. None works when we consider positive values so now consider negative values: - If $x=1$: [1+10 y=20] so, (10 y=-19) thus, (y=-1.overline{9}). - If $x=2$: ( x+10y=-18) thus, (y=-1.overline{8}). - If x=3: ( x+10y=-17) thus, ( y=-1.overline{7}). - If x=4: ( x+10y=-16) thus, ( y=-1.overline{6}). - If x=5: ( x+10y=-15) thus, ( y=-1.overline{5}). - If x=6: ( x+10y=-14) thus, ( y=-1.overline{4}). - If x=7: ( x+10y=-13) thus, ( y=-1.overline{3}). - If x=8: ( x+10y=-12) thus, ( y=-1.overline{2}). Finally, If x=$9$, then, $x+10*y=$11, thus, $11*Y=$11, so, $Y=$$pm$$$dfrac {11}{10}$ so, $Y=$$pm$$$dfrac {11}{9}$ Hence no possible solution exists within given range as well as non-zero constraint. ### Step Summary There are no valid pairs $(x,y)$ satisfying given conditions within given constraints. ### Final Answer(s) json { "x": null, "possible_y": null, "a_20": null } ### Query ### # Self contained Question What does it mean when we say that two angles are complementary? ### Response ### When we say that two angles are complementary, it means that the sum of their measures is exactly 90 degrees. In other words, if you have two angles, let's call