💡First, I need to determine the perimeter of the square garden. Since all sides of a square are equal, the perimeter is calculated by multiplying the side length by 4.Given that the side length is 15 feet, the perimeter would be 4 times 15, which equals 60 feet.Next, I need to find the ratio of the side length to the perimeter. This means comparing the side length of 15 feet to the perimeter of 60 feet.To express this ratio in its simplest form, I divide both numbers by their greatest common divisor, which is 15. Dividing 15 by 15 gives 1, and dividing 60 by 15 gives 4.Therefore, the simplified ratio of the side length to the perimeter is 1 to 4.

💡Okay, so I have this problem here where I need to find a complex number ( z ) that satisfies a certain determinant condition. The determinant is given by the formula:[begin{vmatrix} a & b c & d end{vmatrix} = ad - bc]And the specific condition I need to satisfy is:[begin{vmatrix} z & 1+2i 1-i & 1+i end{vmatrix} = 0]Alright, let me break this down step by step. First, I remember that the determinant of a 2x2 matrix is calculated by multiplying the elements on the main diagonal and subtracting the product of the elements on the other diagonal. So, applying that formula to the given matrix:[begin{vmatrix} z & 1+2i 1-i & 1+i end{vmatrix} = z(1+i) - (1-i)(1+2i)]So, this determinant equals zero. Therefore, I can set up the equation:[z(1+i) - (1-i)(1+2i) = 0]Now, my goal is to solve for ( z ). To do that, I need to isolate ( z ) on one side of the equation. Let me first compute the product ( (1-i)(1+2i) ) because that seems like a necessary step before I can solve for ( z ).Multiplying ( (1 - i) ) and ( (1 + 2i) ):[(1 - i)(1 + 2i) = 1 cdot 1 + 1 cdot 2i - i cdot 1 - i cdot 2i]Calculating each term:- ( 1 cdot 1 = 1 )- ( 1 cdot 2i = 2i )- ( -i cdot 1 = -i )- ( -i cdot 2i = -2i^2 )Now, combining these:[1 + 2i - i - 2i^2]Simplify the like terms:- ( 2i - i = i )- ( -2i^2 ) can be simplified since ( i^2 = -1 ), so ( -2(-1) = 2 )Putting it all together:[1 + i + 2 = 3 + i]So, ( (1 - i)(1 + 2i) = 3 + i )Now, going back to the original equation:[z(1 + i) - (3 + i) = 0]Let me rewrite that:[z(1 + i) = 3 + i]To solve for ( z ), I need to divide both sides by ( (1 + i) ). However, dividing by a complex number can be a bit tricky, so I'll use the conjugate to simplify the expression.The conjugate of ( (1 + i) ) is ( (1 - i) ). I'll multiply both the numerator and the denominator by this conjugate to rationalize the denominator.So,[z = frac{3 + i}{1 + i} times frac{1 - i}{1 - i}]Multiplying the numerators:[(3 + i)(1 - i) = 3 cdot 1 + 3 cdot (-i) + i cdot 1 + i cdot (-i)]Calculating each term:- ( 3 cdot 1 = 3 )- ( 3 cdot (-i) = -3i )- ( i cdot 1 = i )- ( i cdot (-i) = -i^2 )Simplify:[3 - 3i + i - i^2]Combine like terms:- ( -3i + i = -2i )- ( -i^2 = -(-1) = 1 )So, the numerator becomes:[3 - 2i + 1 = 4 - 2i]Now, the denominator:[(1 + i)(1 - i) = 1^2 - (i)^2 = 1 - (-1) = 2]Therefore, putting it all together:[z = frac{4 - 2i}{2} = 2 - i]So, the complex number ( z ) that satisfies the given determinant condition is ( 2 - i ).Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Calculated the determinant correctly: ( z(1 + i) - (1 - i)(1 + 2i) = 0 ).2. Expanded ( (1 - i)(1 + 2i) ) correctly to get ( 3 + i ).3. Set up the equation ( z(1 + i) = 3 + i ).4. Multiplied numerator and denominator by the conjugate ( (1 - i) ) to rationalize.5. Expanded the numerator correctly to ( 4 - 2i ).6. Simplified the denominator to 2.7. Divided to get ( z = 2 - i ).Everything seems to check out. I think my solution is correct.**Final Answer**The complex number ( z ) is boxed{2 - i}.

💡Okay, so I need to find the remainder when the polynomial (5x^3 - 10x^2 + 15x - 20) is divided by (5x - 10). Hmm, I remember something about the Remainder Theorem from my algebra class. Let me try to recall how that works.First, the Remainder Theorem states that if you divide a polynomial (f(x)) by (x - a), the remainder is (f(a)). But in this case, the divisor is (5x - 10), which isn't exactly in the form (x - a). Maybe I can factor out a 5 to make it look more familiar. Let's see:(5x - 10 = 5(x - 2))So, it's the same as (5(x - 2)). That means the divisor is essentially (x - 2) scaled by a factor of 5. I wonder if the Remainder Theorem still applies here. I think it does, but I need to be careful because of the scaling factor.According to the Remainder Theorem, if I divide (f(x)) by (x - a), the remainder is (f(a)). So, if I set (a = 2), then the remainder should be (f(2)). Let me calculate that.Given (f(x) = 5x^3 - 10x^2 + 15x - 20), plugging in (x = 2):(f(2) = 5(2)^3 - 10(2)^2 + 15(2) - 20)Calculating each term step by step:1. (5(2)^3 = 5 times 8 = 40)2. (-10(2)^2 = -10 times 4 = -40)3. (15(2) = 30)4. (-20) remains as is.Now, adding them all together:(40 - 40 + 30 - 20 = 0 + 30 - 20 = 10)So, the remainder is 10. But wait, I factored out a 5 earlier. Does that affect the remainder? I think the scaling factor doesn't change the remainder because the theorem applies to the linear factor (x - a), regardless of the coefficient. So, even though the divisor was (5x - 10), the remainder when dividing by (5x - 10) is still 10.Just to double-check, maybe I can perform polynomial long division to confirm. Let's set it up:Divide (5x^3 - 10x^2 + 15x - 20) by (5x - 10).First term: (5x^3) divided by (5x) is (x^2). Multiply (5x - 10) by (x^2) to get (5x^3 - 10x^2). Subtracting this from the original polynomial:((5x^3 - 10x^2 + 15x - 20) - (5x^3 - 10x^2) = 0x^3 + 0x^2 + 15x - 20)Next term: (15x) divided by (5x) is 3. Multiply (5x - 10) by 3 to get (15x - 30). Subtracting this:((15x - 20) - (15x - 30) = 0x + 10)So, the remainder is indeed 10. That matches what I got using the Remainder Theorem. I think I'm confident now that the remainder is 10.

💡Okay, so I have this problem about a quadrilateral ABCD inscribed in a semicircle with diameter AB. The sides BC and CD are both equal to a, DA is equal to b, and AB is equal to c. All of a, b, c are natural numbers, and I need to show that c cannot be a prime number.Hmm, let me try to visualize this. Since ABCD is inscribed in a semicircle with AB as the diameter, points A and B are at the ends of the diameter, and points C and D lie somewhere on the semicircle. So, ABCD is a cyclic quadrilateral, but it's specifically inscribed in a semicircle, which might give some special properties.I remember that in a cyclic quadrilateral, the opposite angles sum to 180 degrees. But since it's inscribed in a semicircle, maybe some of the angles are right angles? Wait, actually, any triangle inscribed in a semicircle is a right triangle. So, triangles ABC and ABD are right-angled at C and D respectively. That might be useful.Let me write down what I know:- AB = c (diameter)- BC = CD = a- DA = b- All a, b, c are natural numbers- Need to show c is not primeSince ABC and ABD are right-angled triangles, I can apply the Pythagorean theorem to them.For triangle ABC:AC² + BC² = AB²AC² + a² = c²So, AC² = c² - a²Similarly, for triangle ABD:AD² + BD² = AB²b² + BD² = c²So, BD² = c² - b²Now, since ABCD is a quadrilateral, maybe I can apply Ptolemy's theorem. Ptolemy's theorem relates the sides and diagonals of a cyclic quadrilateral. It states that the product of the diagonals is equal to the sum of the products of the opposite sides.So, Ptolemy's theorem for quadrilateral ABCD would be:AC * BD = AB * CD + BC * DASubstituting the known values:AC * BD = c * a + a * bAC * BD = a(c + b)But from earlier, we have expressions for AC² and BD². Let me write those again:AC² = c² - a²BD² = c² - b²So, AC = sqrt(c² - a²) and BD = sqrt(c² - b²)Substituting these into Ptolemy's equation:sqrt(c² - a²) * sqrt(c² - b²) = a(c + b)Hmm, that looks a bit complicated, but maybe I can square both sides to eliminate the square roots.Squaring both sides:(c² - a²)(c² - b²) = a²(c + b)²Let me expand both sides.Left side:(c² - a²)(c² - b²) = c²c² - c²b² - a²c² + a²b² = c⁴ - c²b² - c²a² + a²b²Right side:a²(c + b)² = a²(c² + 2bc + b²) = a²c² + 2a²bc + a²b²So, setting left side equal to right side:c⁴ - c²b² - c²a² + a²b² = a²c² + 2a²bc + a²b²Let me subtract the right side from both sides to bring everything to one side:c⁴ - c²b² - c²a² + a²b² - a²c² - 2a²bc - a²b² = 0Simplify term by term:c⁴ - c²b² - c²a² + a²b² - a²c² - 2a²bc - a²b²Combine like terms:c⁴ - c²b² - c²a² - a²c² - 2a²bc + (a²b² - a²b²)The a²b² terms cancel out:c⁴ - c²b² - 2c²a² - 2a²bc = 0Factor out c² from the first three terms:c²(c² - b² - 2a²) - 2a²bc = 0Hmm, not sure if that helps. Maybe rearrange the equation:c⁴ - c²b² - 2c²a² - 2a²bc = 0Let me factor out c² from the first three terms:c²(c² - b² - 2a²) - 2a²bc = 0Alternatively, maybe factor out c from the first two terms and the last two terms:c²(c² - b²) - 2a²c² - 2a²bc = 0Hmm, not sure. Maybe factor out c from the entire equation:c(c³ - c b² - 2a²c - 2a²b) = 0But c is a natural number, so c ≠ 0, so we have:c³ - c b² - 2a²c - 2a²b = 0Factor out c from the first three terms:c(c² - b² - 2a²) - 2a²b = 0Hmm, not sure. Maybe rearrange terms:c³ - c b² - 2a²c - 2a²b = 0Let me factor terms with c:c³ - c b² - 2a²c = 2a²bFactor c from the left side:c(c² - b² - 2a²) = 2a²bSo, c divides 2a²b. Since a, b, c are natural numbers, c must be a divisor of 2a²b.Now, if c is a prime number, then c must divide one of the factors in 2a²b. Since c divides 2a²b, and c is prime, c divides 2, c divides a, or c divides b.But c is the diameter of the semicircle, which is the longest side in the quadrilateral. So, c must be greater than a and b because in a semicircle, the diameter is the longest chord.Therefore, c > a and c > b. So, c cannot divide a or b because c is larger than both. The only possibility is that c divides 2.But c is a natural number, so c can be 1 or 2. But c is the diameter, which must be at least longer than the radius, so c must be at least 2.If c = 2, let's see if that's possible.If c = 2, then from the earlier equation:c(c² - b² - 2a²) = 2a²bSubstitute c = 2:2(4 - b² - 2a²) = 2a²bDivide both sides by 2:4 - b² - 2a² = a²bRearrange:4 = a²b + b² + 2a²Hmm, let's see if there are natural numbers a and b such that this holds.Let me try a = 1:4 = 1*b + b² + 2*1 = b + b² + 2So, b² + b + 2 = 4b² + b - 2 = 0Solutions: b = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2So, b = 1 or b = -2. Since b is natural, b = 1.So, a = 1, b = 1, c = 2.But in the problem, it's given that b ≠ a, so b = 1 and a = 1 would violate that. So, a cannot be 1.Next, try a = 2:4 = (2)^2 * b + b² + 2*(2)^2 = 4b + b² + 8So, 4 = 4b + b² + 8Rearrange:b² + 4b + 8 - 4 = 0b² + 4b + 4 = 0This factors as (b + 2)^2 = 0, so b = -2, which is not natural.So, a = 2 doesn't work.a = 3:4 = 9b + b² + 18So, 4 = b² + 9b + 18Rearrange:b² + 9b + 14 = 0Discriminant: 81 - 56 = 25Solutions: b = [-9 ± 5]/2So, b = (-9 + 5)/2 = -2 or b = (-9 -5)/2 = -7. Not natural.a = 4:4 = 16b + b² + 32So, 4 = b² + 16b + 32Rearrange:b² + 16b + 28 = 0Discriminant: 256 - 112 = 144Solutions: b = [-16 ± 12]/2b = (-16 + 12)/2 = -2 or b = (-16 -12)/2 = -14. Not natural.Hmm, seems like for c = 2, the only possible solution is a = 1, b = 1, which violates b ≠ a. So, c = 2 is not possible.Therefore, c cannot be 2, and since c must divide 2a²b, and c is prime, the only possible prime divisor is 2, which doesn't work. Therefore, c cannot be prime.Wait, let me double-check. If c is a prime number, it must divide 2a²b. Since c > a and c > b, c cannot divide a or b, so c must divide 2. Therefore, c must be 2, but we saw that c = 2 doesn't work because it leads to a = b, which is not allowed. Hence, c cannot be prime.Yeah, that seems to make sense. So, c cannot be a prime number.**Final Answer**boxed{c} cannot be a prime number.

💡Okay, so I have this problem here about a triangle where one of the angles is given as 180 degrees minus n degrees. The difference between the largest and smallest angles in the triangle is 24 degrees. I need to find the range of values for n. Hmm, let me try to figure this out step by step.First, I know that in any triangle, the sum of the three angles is always 180 degrees. So, if one angle is 180 - n, the other two angles must add up to n degrees. Let me denote the other two angles as angle A and angle B. So, angle A + angle B = n degrees.Now, the problem also says that the difference between the largest and smallest angles is 24 degrees. That means if I arrange the three angles in order, the biggest one minus the smallest one is 24 degrees. So, I need to consider different cases depending on whether 180 - n is the largest angle, the smallest angle, or somewhere in the middle.Let me think about the first case: suppose 180 - n is neither the largest nor the smallest angle. That means angle A and angle B are such that one is the largest and the other is the smallest, with a difference of 24 degrees. So, without loss of generality, let's say angle A is the smallest and angle B is the largest. Therefore, angle B - angle A = 24 degrees.Since angle A + angle B = n, I can write angle B = angle A + 24. Substituting that into the sum, angle A + (angle A + 24) = n, which simplifies to 2 angle A + 24 = n. So, angle A = (n - 24)/2.Now, I also know that all angles in a triangle must be positive and less than 180 degrees. So, angle A must be greater than 0, which gives (n - 24)/2 > 0, so n > 24. Similarly, angle B must be less than 180, but since angle B = angle A + 24, and angle A is (n - 24)/2, angle B = (n - 24)/2 + 24 = (n + 24)/2. So, (n + 24)/2 < 180, which gives n + 24 < 360, so n < 336. But since n is part of an angle in a triangle, it's more constrained.Wait, actually, 180 - n is one of the angles, so 180 - n must also be positive, so 180 - n > 0, which means n < 180. So, n has to be less than 180. So, combining these, n must be greater than 24 and less than 180.But we also have to make sure that 180 - n is neither the largest nor the smallest angle. So, 180 - n must be greater than angle A and less than angle B. So, 180 - n > angle A and 180 - n < angle B.We already have angle A = (n - 24)/2 and angle B = (n + 24)/2. So, 180 - n > (n - 24)/2 and 180 - n < (n + 24)/2.Let me solve the first inequality: 180 - n > (n - 24)/2.Multiply both sides by 2: 360 - 2n > n - 24.Bring all terms to one side: 360 + 24 > 3n.So, 384 > 3n, which simplifies to n < 128.Now, the second inequality: 180 - n < (n + 24)/2.Multiply both sides by 2: 360 - 2n < n + 24.Bring all terms to one side: 360 - 24 < 3n.So, 336 < 3n, which simplifies to n > 112.So, in this case, n must be greater than 112 and less than 128.Okay, that's the first case where 180 - n is neither the largest nor the smallest angle.Now, let's consider the second case: 180 - n is the largest angle. That means the difference between 180 - n and the smallest angle is 24 degrees. So, the smallest angle is 180 - n - 24 = 156 - n degrees.Since the sum of all angles is 180, the third angle would be 180 - (180 - n) - (156 - n) = 180 - 180 + n - 156 + n = 2n - 156 degrees.So, the three angles are 180 - n, 156 - n, and 2n - 156.Now, all angles must be positive. So, 156 - n > 0, which gives n < 156. Also, 2n - 156 > 0, which gives n > 78.Additionally, since 180 - n is the largest angle, it must be greater than the other two angles. So, 180 - n > 156 - n, which is always true because 180 > 156. Also, 180 - n > 2n - 156.Let me solve that inequality: 180 - n > 2n - 156.Bring all terms to one side: 180 + 156 > 3n.So, 336 > 3n, which simplifies to n < 112.But earlier, we had n > 78 and n < 156. So, combining these, n must be greater than 78 and less than 112.Wait, but we also have that 156 - n must be the smallest angle, so 156 - n < 2n - 156.Let me solve that: 156 - n < 2n - 156.Bring all terms to one side: 156 + 156 < 3n.So, 312 < 3n, which simplifies to n > 104.So, combining all these, n must be greater than 104 and less than 112.Okay, that's the second case where 180 - n is the largest angle.Now, the third case: 180 - n is the smallest angle. That means the difference between the largest angle and 180 - n is 24 degrees. So, the largest angle is 180 - n + 24 = 204 - n degrees.The third angle would be 180 - (180 - n) - (204 - n) = 180 - 180 + n - 204 + n = 2n - 204 degrees.So, the three angles are 180 - n, 204 - n, and 2n - 204.Again, all angles must be positive. So, 204 - n > 0, which gives n < 204. Also, 2n - 204 > 0, which gives n > 102.Additionally, since 180 - n is the smallest angle, it must be less than the other two angles. So, 180 - n < 204 - n, which is always true because 180 < 204. Also, 180 - n < 2n - 204.Let me solve that inequality: 180 - n < 2n - 204.Bring all terms to one side: 180 + 204 < 3n.So, 384 < 3n, which simplifies to n > 128.But earlier, we had n > 102 and n < 204. So, combining these, n must be greater than 128 and less than 204.Wait, but we also have that 204 - n must be the largest angle, so 204 - n > 2n - 204.Let me solve that: 204 - n > 2n - 204.Bring all terms to one side: 204 + 204 > 3n.So, 408 > 3n, which simplifies to n < 136.So, combining all these, n must be greater than 128 and less than 136.Okay, that's the third case where 180 - n is the smallest angle.Now, let me summarize the three cases:1. When 180 - n is neither the largest nor the smallest angle: 112 < n < 128.2. When 180 - n is the largest angle: 104 < n < 112.3. When 180 - n is the smallest angle: 128 < n < 136.So, combining all these ranges, n can be from 104 to 136 degrees. But I need to make sure there are no overlaps or gaps.Looking at the ranges:- 104 < n < 112- 112 < n < 128- 128 < n < 136These are consecutive and cover from 104 to 136 without gaps. So, the overall range is 104 ≤ n ≤ 136.Wait, but in each case, the inequalities were strict (greater than or less than), but since n can take the boundary values where the angle differences become exactly 24 degrees, I think n can be equal to 104, 112, 128, and 136 as well.Let me check n = 104:If n = 104, then 180 - n = 76 degrees. The other angles would be 156 - n = 52 degrees and 2n - 156 = 208 - 156 = 52 degrees. Wait, that would make two angles equal to 52 degrees, and the third angle 76 degrees. The difference between the largest (76) and smallest (52) is 24 degrees. So, that works.Similarly, n = 112:180 - 112 = 68 degrees. The other angles would be 156 - 112 = 44 degrees and 2*112 - 156 = 224 - 156 = 68 degrees. So, angles are 68, 44, and 68. The difference between largest (68) and smallest (44) is 24 degrees. That works.n = 128:180 - 128 = 52 degrees. The other angles would be 204 - 128 = 76 degrees and 2*128 - 204 = 256 - 204 = 52 degrees. So, angles are 52, 76, and 52. The difference between largest (76) and smallest (52) is 24 degrees. That works.n = 136:180 - 136 = 44 degrees. The other angles would be 204 - 136 = 68 degrees and 2*136 - 204 = 272 - 204 = 68 degrees. So, angles are 44, 68, and 68. The difference between largest (68) and smallest (44) is 24 degrees. That works.So, including the endpoints, the range is 104 ≤ n ≤ 136.Therefore, the range of values for n is from 104 degrees to 136 degrees.

💡First, I need to evaluate the cosine of ( frac{7pi}{4} ). I recall that ( frac{7pi}{4} ) is in the fourth quadrant of the unit circle, where the cosine values are positive.Next, I recognize that ( frac{7pi}{4} ) is equivalent to ( 2pi - frac{pi}{4} ). This means that ( cosleft(frac{7pi}{4}right) ) is the same as ( cosleft(frac{pi}{4}right) ).I know that ( cosleft(frac{pi}{4}right) ) equals ( frac{sqrt{2}}{2} ).Now, I need to calculate the logarithm base 2 of ( frac{sqrt{2}}{2} ). I can rewrite ( frac{sqrt{2}}{2} ) as ( 2^{-frac{1}{2}} ).Finally, applying the logarithm property ( log_{2}(2^{x}) = x ), I find that ( log_{2}(2^{-frac{1}{2}}) = -frac{1}{2} ).Therefore, the value of ( log_{2}left(cosleft(frac{7pi}{4}right)right) ) is ( -frac{1}{2} ).