💡Okay, so I have this problem about two rectangles. The perimeter of one is four times the perimeter of the other, and I need to figure out how many times the area of the larger rectangle is compared to the smaller one. They also mention that the sides are proportional, which probably means they're similar rectangles.Alright, let's start by recalling some basic formulas. The perimeter of a rectangle is calculated as (2 times (length + width)). The area is (length times width). Since the rectangles are proportional, their sides must be in the same ratio. Let's call the sides of the smaller rectangle (l) and (w), and the sides of the larger rectangle (kl) and (kw), where (k) is the scale factor.Given that the perimeter of the larger rectangle is four times that of the smaller one, I can set up an equation. The perimeter of the smaller rectangle is (2(l + w)), and the perimeter of the larger one is (2(kl + kw)). According to the problem, the larger perimeter is four times the smaller one:[2(kl + kw) = 4 times 2(l + w)]Simplifying this, I get:[2k(l + w) = 8(l + w)]Dividing both sides by (2(l + w)) (assuming (l + w neq 0)):[k = 4]So the scale factor (k) is 4. That means each side of the larger rectangle is four times the corresponding side of the smaller rectangle.Now, let's look at the areas. The area of the smaller rectangle is (l times w), and the area of the larger rectangle is (kl times kw = k^2 times l times w). Substituting (k = 4):[Area_{text{larger}} = 4^2 times Area_{text{smaller}} = 16 times Area_{text{smaller}}]So, the area of the larger rectangle is 16 times the area of the smaller rectangle.Wait, let me double-check that. If the scale factor for the sides is 4, then the scale factor for the area should be (4^2 = 16). Yeah, that makes sense because area is a two-dimensional measurement, so it scales with the square of the linear dimensions.Therefore, the answer should be 16.

💡Okay, so I've got this problem where four people—Person A, Person B, Person C, and Person D—are sharing 2013 candies. The problem gives me some relationships between how many candies each person gets. Specifically, it says:- Person A gets 10 more candies than twice the amount Person B has.- Person A gets 18 more candies than three times the amount Person C has.- Person A gets 55 less candies than five times the amount Person D has.And the question is asking how many candies Person A gets. Alright, let's break this down step by step.First, I think it would help to assign variables to each person to make the relationships clearer. Let me denote:- Let ( A ) be the number of candies Person A gets.- Let ( B ) be the number of candies Person B gets.- Let ( C ) be the number of candies Person C gets.- Let ( D ) be the number of candies Person D gets.Now, according to the problem:1. Person A gets 10 more candies than twice the amount Person B has. So, mathematically, that translates to: [ A = 2B + 10 ] 2. Person A gets 18 more candies than three times the amount Person C has. So: [ A = 3C + 18 ] 3. Person A gets 55 less candies than five times the amount Person D has. So: [ A = 5D - 55 ] Additionally, we know that the total number of candies they all share is 2013. So: [ A + B + C + D = 2013 ]Alright, so now I have four equations:1. ( A = 2B + 10 )2. ( A = 3C + 18 )3. ( A = 5D - 55 )4. ( A + B + C + D = 2013 )My goal is to find the value of ( A ). To do that, I can express ( B ), ( C ), and ( D ) in terms of ( A ) using the first three equations and then substitute them into the fourth equation.Starting with the first equation:[ A = 2B + 10 ]I can solve for ( B ):[ 2B = A - 10 ][ B = frac{A - 10}{2} ]Similarly, from the second equation:[ A = 3C + 18 ]Solving for ( C ):[ 3C = A - 18 ][ C = frac{A - 18}{3} ]From the third equation:[ A = 5D - 55 ]Solving for ( D ):[ 5D = A + 55 ][ D = frac{A + 55}{5} ]Now, I have expressions for ( B ), ( C ), and ( D ) in terms of ( A ). Let's plug these into the fourth equation:[ A + B + C + D = 2013 ]Substituting the expressions we found:[ A + frac{A - 10}{2} + frac{A - 18}{3} + frac{A + 55}{5} = 2013 ]Hmm, this looks a bit complicated with all these fractions. Maybe I can find a common denominator to combine them. The denominators are 2, 3, and 5. The least common multiple of these is 30. So, I'll multiply each term by 30 to eliminate the denominators.Multiplying each term by 30:[ 30A + 15(A - 10) + 10(A - 18) + 6(A + 55) = 30 times 2013 ]Let me compute each part step by step.First, expand each term:- ( 30A ) stays as it is.- ( 15(A - 10) = 15A - 150 )- ( 10(A - 18) = 10A - 180 )- ( 6(A + 55) = 6A + 330 )Now, combine all these:[ 30A + 15A - 150 + 10A - 180 + 6A + 330 = 60390 ]Combine like terms:- Combine the ( A ) terms: ( 30A + 15A + 10A + 6A = 61A )- Combine the constant terms: ( -150 - 180 + 330 = 0 )Wait, that's interesting. The constants cancel out:[ -150 - 180 + 330 = (-150 - 180) + 330 = (-330) + 330 = 0 ]So, the equation simplifies to:[ 61A = 60390 ]Now, solve for ( A ):[ A = frac{60390}{61} ]Let me compute that. Dividing 60390 by 61.First, see how many times 61 goes into 60390.61 times 990 is 60390 because 61 times 1000 is 61000, which is 610 more than 60390. So, 61000 - 610 = 60390, which means 61 times 990 is 60390.Therefore, ( A = 990 ).Let me double-check my calculations to make sure I didn't make a mistake.First, the expressions for ( B ), ( C ), and ( D ) in terms of ( A ):- ( B = frac{A - 10}{2} = frac{990 - 10}{2} = frac{980}{2} = 490 )- ( C = frac{A - 18}{3} = frac{990 - 18}{3} = frac{972}{3} = 324 )- ( D = frac{A + 55}{5} = frac{990 + 55}{5} = frac{1045}{5} = 209 )Now, let's add them up:- ( A = 990 )- ( B = 490 )- ( C = 324 )- ( D = 209 )Total: ( 990 + 490 = 1480 ); ( 1480 + 324 = 1804 ); ( 1804 + 209 = 2013 )Yes, that adds up to 2013, which matches the total number of candies. So, my calculations seem correct.Therefore, Person A gets 990 candies.

💡First, I need to determine the area of the outermost circle with a radius of 15 units. Using the formula for the area of a circle, ( pi r^2 ), the area is ( pi times 15^2 = 225pi ).Next, I calculate the area of the innermost circle with a radius of 4 units. Applying the same formula, the area is ( pi times 4^2 = 16pi ).Finally, to find the area of the ring-shaped region between the innermost and outermost circles, I subtract the area of the innermost circle from the area of the outermost circle: ( 225pi - 16pi = 209pi ).

💡First, I need to determine all possible sums Tamika can obtain by adding two different numbers from the set {7, 10, 12}. - 7 + 10 = 17- 7 + 12 = 19- 10 + 12 = 22Next, I calculate all possible products Carlos can obtain by multiplying two different numbers from the set {-4, 2, 6}.- (-4) × 2 = -8- (-4) × 6 = -24- 2 × 6 = 12Now, I compare each of Tamika's sums with each of Carlos' products to see when Tamika's result is greater than Carlos' result.- 17 > -8 (True)- 17 > -24 (True)- 17 > 12 (True)- 19 > -8 (True)- 19 > -24 (True)- 19 > 12 (True)- 22 > -8 (True)- 22 > -24 (True)- 22 > 12 (True)Out of the 9 possible comparisons, Tamika's result is greater in 8 instances. Therefore, the probability that Tamika's result is greater than Carlos' result is 8/9.

💡First, I need to determine the total number of each type of coin Rocco has by multiplying the number of piles by the number of coins in each pile.Next, I'll calculate the total value of each type of coin in their respective currencies:- For U.S. Quarters and Nickels, the value is already in U.S. Dollars.- For Canadian Dimes, I'll convert the total value from Canadian Dollars to U.S. Dollars using the given exchange rate.- For Euro Cents, I'll convert the total value from Euros to U.S. Dollars using the given exchange rate.- For British Pence, I'll convert the total value from British Pounds to U.S. Dollars using the given exchange rate.After converting all amounts to U.S. Dollars, I'll sum them up to find the total amount of money Rocco has in U.S. Dollars.

💡Okay, so I have these two math problems here, and I need to figure them out step by step. Let me start with the first one.**Problem 1:**Given that tan(π/4 + α) = 2, find the value of sin2α + cos²α.Alright, I remember that tan(A + B) has a formula. Let me recall it. I think it's tan(A + B) = (tanA + tanB) / (1 - tanA tanB). So, in this case, A is π/4 and B is α. Since tan(π/4) is 1, right? So substituting that in, tan(π/4 + α) would be (1 + tanα) / (1 - tanα). And this is equal to 2.So, I can set up the equation: (1 + tanα) / (1 - tanα) = 2. Let me solve for tanα.Multiply both sides by (1 - tanα): 1 + tanα = 2(1 - tanα). Expanding the right side: 1 + tanα = 2 - 2 tanα. Now, let's get all the tanα terms on one side and constants on the other.Add 2 tanα to both sides: 1 + 3 tanα = 2. Then subtract 1 from both sides: 3 tanα = 1. So, tanα = 1/3.Okay, so tanα is 1/3. Now, I need to find sin2α + cos²α. Hmm, I remember that sin2α is 2 sinα cosα. So, sin2α + cos²α = 2 sinα cosα + cos²α.Maybe I can factor out cosα: cosα (2 sinα + cosα). But I'm not sure if that helps. Alternatively, maybe I can express everything in terms of tanα since I know tanα = 1/3.Let me think. If tanα = 1/3, then I can imagine a right triangle where the opposite side is 1 and the adjacent side is 3. Then, the hypotenuse would be sqrt(1² + 3²) = sqrt(10). So, sinα = opposite/hypotenuse = 1/sqrt(10), and cosα = adjacent/hypotenuse = 3/sqrt(10).So, sinα = 1/sqrt(10), cosα = 3/sqrt(10). Let me compute sin2α + cos²α.First, sin2α = 2 sinα cosα = 2*(1/sqrt(10))*(3/sqrt(10)) = 6/10 = 3/5.Next, cos²α = (3/sqrt(10))² = 9/10.So, sin2α + cos²α = 3/5 + 9/10. To add these, I need a common denominator, which is 10.3/5 is 6/10, so 6/10 + 9/10 = 15/10 = 3/2.So, the value is 3/2. That seems straightforward.Wait, let me double-check. Maybe there's another way to compute this without going through the triangle. I remember that sin2α + cos²α can be expressed in terms of tanα.Let me try that. Since tanα = 1/3, I can write sin2α as 2 tanα / (1 + tan²α). And cos²α is 1 / (1 + tan²α).So, sin2α + cos²α = (2 tanα / (1 + tan²α)) + (1 / (1 + tan²α)) = (2 tanα + 1) / (1 + tan²α).Substituting tanα = 1/3: (2*(1/3) + 1) / (1 + (1/3)²) = (2/3 + 1) / (1 + 1/9) = (5/3) / (10/9) = (5/3)*(9/10) = (45/30) = 3/2.Same result. Okay, so that checks out. So, I'm confident that the answer is 3/2.**Problem 2:**Points A(x₁, y₁) and B(x₂, y₂) lie on a unit circle centered at the origin O. The minor arc AB corresponds to a central angle α. If sinα + cosα = 7/17, find the value of x₁x₂ + y₁y₂.Alright, so points A and B are on the unit circle, which means their coordinates satisfy x₁² + y₁² = 1 and x₂² + y₂² = 1.The minor arc AB corresponds to a central angle α, which means the angle between vectors OA and OB is α. So, the dot product of vectors OA and OB is equal to |OA||OB|cosα. Since it's a unit circle, |OA| = |OB| = 1, so the dot product is just cosα.But the dot product is also equal to x₁x₂ + y₁y₂. So, x₁x₂ + y₁y₂ = cosα.Therefore, if I can find cosα, I can find the value of x₁x₂ + y₁y₂.Given that sinα + cosα = 7/17. Let me denote this as equation (1).I need to find cosα. Maybe I can square both sides of equation (1) to get an expression involving sinα cosα.So, (sinα + cosα)² = (7/17)².Expanding the left side: sin²α + 2 sinα cosα + cos²α = 49/289.But sin²α + cos²α = 1, so 1 + 2 sinα cosα = 49/289.Therefore, 2 sinα cosα = 49/289 - 1 = 49/289 - 289/289 = -240/289.So, sinα cosα = -120/289.Hmm, okay. Now, I have sinα + cosα = 7/17 and sinα cosα = -120/289.I need to find cosα. Let me think about solving these equations.Let me denote sinα = s and cosα = c. Then, s + c = 7/17 and s*c = -120/289.I can set up a quadratic equation where the roots are s and c. The quadratic would be t² - (s + c)t + s*c = 0, which is t² - (7/17)t - 120/289 = 0.Let me write that as: t² - (7/17)t - 120/289 = 0.Multiply both sides by 289 to eliminate denominators: 289 t² - 119 t - 120 = 0.Now, let's solve for t using quadratic formula.t = [119 ± sqrt(119² + 4*289*120)] / (2*289).First, compute discriminant D: 119² + 4*289*120.119²: 119*119. Let me compute that. 120² is 14400, so 119² is 14400 - 2*120 + 1 = 14400 - 240 + 1 = 14161.4*289*120: 4*289 = 1156; 1156*120. Let me compute 1156*100 = 115600, 1156*20=23120, so total is 115600 + 23120 = 138720.So, D = 14161 + 138720 = 152881.Now, sqrt(152881). Let me see, 391² is 152881 because 400² is 160000, so 390² is 152100, 391² = 152100 + 2*390 +1 = 152100 + 780 +1=152881. So sqrt(D)=391.Therefore, t = [119 ± 391]/(2*289).Compute both roots:First root: (119 + 391)/578 = 510/578. Simplify: divide numerator and denominator by 34: 510 ÷34=15, 578 ÷34=17. So, 15/17.Second root: (119 - 391)/578 = (-272)/578. Simplify: divide numerator and denominator by 34: -272 ÷34= -8, 578 ÷34=17. So, -8/17.So, the solutions are t = 15/17 and t = -8/17.So, sinα and cosα are 15/17 and -8/17, but which one is which?We know that sinα + cosα =7/17. Let's check:If sinα =15/17 and cosα = -8/17, then sinα + cosα =15/17 -8/17=7/17. That works.Alternatively, if sinα =-8/17 and cosα=15/17, then sinα + cosα= -8/17 +15/17=7/17. That also works.But wait, we need to figure out which is which. Let me think about the angle α.Given that points A and B are on the unit circle, and the minor arc AB corresponds to central angle α. So, α is between 0 and π, because it's a minor arc.So, α is in (0, π). Therefore, cosα can be positive or negative, but sinα is positive because in (0, π), sine is positive.So, if sinα is positive, then sinα must be 15/17, and cosα must be -8/17.Therefore, cosα = -8/17.Therefore, x₁x₂ + y₁y₂ = cosα = -8/17.So, the value is -8/17.Wait, let me double-check.We had sinα + cosα=7/17, and sinα cosα=-120/289.We found sinα=15/17, cosα=-8/17.Check sinα + cosα=15/17 -8/17=7/17. Correct.Check sinα cosα=15/17*(-8)/17=-120/289. Correct.So, that's consistent.Therefore, x₁x₂ + y₁y₂=cosα=-8/17.So, that's the answer.**Final Answer**1. boxed{dfrac{3}{2}}2. boxed{-dfrac{8}{17}}