answer：Okay, so I need to find the determinant of this 3x3 matrix:[begin{vmatrix} 3 & 0 & 1 -5 & 5 & -4 3 & 3 & 6 end{vmatrix}]Hmm, I remember that to calculate the determinant of a 3x3 matrix, one way is to use the method of expansion by minors. I think you pick a row or column to expand along, and then for each element in that row or column, you multiply it by its corresponding minor and apply the appropriate sign based on its position.I think the first step is to choose a row or column to expand. Usually, people choose the row or column with the most zeros because that can make the calculations easier. Looking at this matrix, the first row has a zero in the second position. That might be a good choice because it can simplify the calculation. Let me go with the first row.So, the determinant can be expanded along the first row as follows:[text{det} = 3 times M_{11} - 0 times M_{12} + 1 times M_{13}]Where ( M_{11} ), ( M_{12} ), and ( M_{13} ) are the minors corresponding to the elements in the first row.Since the second term has a zero, it will vanish, so I only need to calculate the first and the third terms.First, let's find the minor ( M_{11} ). The minor is the determinant of the 2x2 matrix that remains after removing the first row and the first column. So, removing the first row and first column, we get:[begin{vmatrix} 5 & -4 3 & 6 end{vmatrix}]To calculate this determinant, I multiply the elements on the main diagonal and subtract the product of the elements on the secondary diagonal:[M_{11} = (5)(6) - (-4)(3) = 30 - (-12) = 30 + 12 = 42]Okay, so ( M_{11} = 42 ).Next, let's find the minor ( M_{13} ). This is the determinant of the 2x2 matrix that remains after removing the first row and the third column. Removing the first row and third column, we get:[begin{vmatrix} -5 & 5 3 & 3 end{vmatrix}]Again, calculating this determinant:[M_{13} = (-5)(3) - (5)(3) = -15 - 15 = -30]So, ( M_{13} = -30 ).Now, plugging these back into the expansion formula:[text{det} = 3 times 42 - 0 times M_{12} + 1 times (-30)]Simplifying this:[text{det} = 126 + 0 - 30 = 96]Wait, that seems straightforward. But just to make sure I didn't make any mistakes, let me double-check my calculations.First minor ( M_{11} ):[begin{vmatrix} 5 & -4 3 & 6 end{vmatrix} = (5)(6) - (-4)(3) = 30 + 12 = 42]Yes, that's correct.Second minor ( M_{13} ):[begin{vmatrix} -5 & 5 3 & 3 end{vmatrix} = (-5)(3) - (5)(3) = -15 - 15 = -30]That also looks correct.So, multiplying and adding:[3 times 42 = 126][1 times (-30) = -30][126 - 30 = 96]Yes, that seems right. I don't see any errors in my steps. Therefore, the determinant of the given matrix is 96.

answer：Alright, so I've got this problem here where I need to figure out the relationship between three numbers: a, b, and c. Let me write them down again to make sure I have them right.a is equal to 0.5 raised to the power of 0.5. Okay, so that's like the square root of 0.5, right? Because 0.5 is the same as 1/2, so the square root of 1/2. Hmm, I remember that the square root of 1/2 is approximately 0.707. Let me double-check that. Yeah, sqrt(1/2) is about 0.707. So a is roughly 0.707.Next, b is 0.3 raised to the power of 0.5. That's the square root of 0.3. I think the square root of 0.3 is a bit less than the square root of 0.25, which is 0.5, so maybe around 0.547? Let me calculate it more accurately. 0.547 squared is about 0.299, which is close to 0.3, so yeah, b is approximately 0.547.Now, c is a logarithm: log base 0.3 of 0.2. Hmm, logarithms can be tricky. I remember that log base a of b is the exponent you need to raise a to get b. So, log_{0.3}(0.2) is the number x such that 0.3^x = 0.2. Since 0.3 is less than 1, the logarithm function is decreasing, right? So, as the input increases, the output decreases.Let me think about this. If I have log_{0.3}(0.3), that's 1 because 0.3^1 = 0.3. And log_{0.3}(1) would be 0 because 0.3^0 = 1. But 0.2 is less than 0.3, so what does that mean for the logarithm? Since the function is decreasing, a smaller input gives a larger output. So, log_{0.3}(0.2) should be greater than 1.Wait, let me verify that. If 0.3^1 = 0.3, and 0.3^x = 0.2, which is smaller than 0.3, so x has to be greater than 1 because as x increases, 0.3^x decreases. So, c is greater than 1.Okay, so now I have approximate values:a ≈ 0.707b ≈ 0.547c > 1So, comparing these, b is the smallest, then a, then c. So the order from smallest to largest is b < a < c. Looking at the options, that's option B: b < a < c.Wait, let me make sure I didn't mix up anything. The function y = x^0.5 is increasing, so since 0.5 > 0.3, then 0.5^0.5 > 0.3^0.5, which means a > b. That makes sense. And for c, since log_{0.3}(0.2) > 1, and 1 is greater than both a and b, so c is the largest. Yeah, that seems right.I think I got it. So, the correct relationship is b < a < c, which is option B.

answer：Okay, so I'm trying to solve this geometry problem. Let me read it again and make sure I understand all the given information.We have a diagram where point P is on line RS. There are two angle bisectors: QP bisects angle SQR, and RP bisects angle PRS. Also, it's given that PQ equals PR. The measure of angle RSQ is 4x degrees, and the measure of angle QRP is 5x degrees. We need to find the measure of angle RQP.Hmm, let me try to visualize the diagram. Points R, S, and Q form a triangle, I assume. Point P is somewhere on RS. QP is an angle bisector of angle SQR, which means it splits angle SQR into two equal parts. Similarly, RP is an angle bisector of angle PRS, so it splits angle PRS into two equal parts.Also, PQ equals PR, which tells me that triangle PQR is isosceles with PQ and PR as the equal sides. Therefore, the base angles at Q and R should be equal. Wait, but angle RQP is one of those base angles, right? So maybe that will help me later.Given angles: angle RSQ is 4x degrees. That's the angle at point S in triangle RSQ. Also, angle QRP is 5x degrees. That's the angle at point R in triangle QRP.I think I need to label all the given angles and use the properties of angle bisectors and isosceles triangles to find the value of x first, and then compute angle RQP.Let me start by drawing triangle RSQ. Point P is on RS, so I'll place P somewhere between R and S. QP is the angle bisector of angle SQR, so it splits angle SQR into two equal angles. Similarly, RP is the angle bisector of angle PRS, splitting it into two equal angles.Since PQ equals PR, triangle PQR is isosceles with PQ = PR. Therefore, angles opposite these sides are equal. That means angle PQR (which is angle RQP) is equal to angle PRQ. Wait, is that correct? Let me think.In triangle PQR, sides PQ and PR are equal, so the angles opposite them, which are angles PRQ and PQR, should be equal. So angle PRQ equals angle PQR. But angle PQR is the same as angle RQP because they are the same angle. Wait, no, angle PQR is at point Q, and angle RQP is also at point Q. Hmm, maybe I'm confusing the notation.Wait, angle RQP is the angle at point Q between sides RQ and PQ. So in triangle PQR, angle RQP is at point Q, and since PQ = PR, the base angles at Q and R are equal. So angle RQP equals angle PRQ. That makes sense.So, angle RQP = angle PRQ.Now, let's denote some variables. Let me assign angle SQR as 2θ because QP is its bisector, so each of the two angles created by the bisector will be θ. Similarly, since RP bisects angle PRS, let me denote angle PRS as 2φ, so each of the angles created by the bisector RP will be φ.Given that angle RSQ is 4x degrees. That's the angle at point S in triangle RSQ. So in triangle RSQ, we have angles at R, S, and Q. Let me denote angle RSQ as 4x, angle SQR as 2θ, and angle QRS as something else.Wait, angle QRS is at point R in triangle RSQ. But point P is on RS, so angle QRS is the same as angle QRP in triangle QRP, right? Wait, no, angle QRP is given as 5x degrees. So angle QRP is at point R in triangle QRP, which is part of triangle RSQ.Wait, maybe I need to consider triangle RSQ and triangle QRP separately.In triangle RSQ, we have:- Angle at S: angle RSQ = 4x- Angle at Q: angle SQR = 2θ- Angle at R: angle QRS = ?Since the sum of angles in a triangle is 180 degrees, we have:angle RSQ + angle SQR + angle QRS = 1804x + 2θ + angle QRS = 180So, angle QRS = 180 - 4x - 2θBut angle QRS is also angle QRP in triangle QRP, which is given as 5x degrees. Wait, is that correct?Wait, point P is on RS, so angle QRS is the same as angle QRP because they are both angles at point R between lines RQ and RS. So angle QRS = angle QRP = 5x degrees.Therefore, from the earlier equation:angle QRS = 180 - 4x - 2θ = 5xSo, 180 - 4x - 2θ = 5xSimplify: 180 - 2θ = 9xSo, 2θ = 180 - 9xTherefore, θ = (180 - 9x)/2Okay, so θ is expressed in terms of x.Now, let's look at triangle PQR. In triangle PQR, we have:- PQ = PR, so it's isosceles with base angles at Q and R equal. Therefore, angle RQP = angle PRQ.We also know that QP bisects angle SQR, which we denoted as 2θ, so each of the angles created by the bisector is θ. Therefore, angle PQS = angle PQR = θ.Wait, angle PQR is at point Q, so angle PQR is θ, and since triangle PQR is isosceles with PQ = PR, angle PRQ is also θ. Therefore, angle RQP = θ.Wait, but earlier I thought angle RQP = angle PRQ, which is θ. So angle RQP is θ.But we need to find angle RQP in terms of x. So if we can find θ in terms of x, we can find angle RQP.From earlier, θ = (180 - 9x)/2But let's see if we can find another equation involving θ and φ.Since RP bisects angle PRS, which we denoted as 2φ, so each of the angles created by the bisector is φ. Therefore, angle PRS = angle PRQ = φ.Wait, angle PRS is at point R in triangle PRS, and angle PRQ is at point R in triangle PQR. Are these the same angles?Wait, point P is on RS, so angle PRS is the angle at point R between lines PR and RS, and angle PRQ is the angle at point R between lines PR and RQ. So they are different angles.Wait, maybe I need to clarify.In triangle PRS, RP bisects angle PRS, so angle PRS is 2φ, and each of the angles created by the bisector is φ. So angle PRS is split into two angles of φ each.Similarly, in triangle PQR, since PQ = PR, angles at Q and R are equal, which are angle RQP and angle PRQ, both equal to θ.But angle PRQ is part of triangle PQR, while angle PRS is part of triangle PRS.Wait, perhaps I need to consider triangle PRS.In triangle PRS, we have:- Angle at P: angle RPS- Angle at R: angle PRS = 2φ- Angle at S: angle PSRSince the sum of angles in a triangle is 180 degrees, we have:angle RPS + angle PRS + angle PSR = 180angle RPS + 2φ + angle PSR = 180But I don't know angle RPS or angle PSR yet.Wait, angle RPS is part of triangle PQR as well. Let me see.In triangle PQR, we have angles:- At Q: angle RQP = θ- At R: angle PRQ = θ- At P: angle QPRSum of angles: θ + θ + angle QPR = 180So, angle QPR = 180 - 2θBut angle QPR is also part of triangle PRS. In triangle PRS, angle QPR is the same as angle RPS. So angle RPS = 180 - 2θTherefore, in triangle PRS:angle RPS = 180 - 2θangle PRS = 2φangle PSR = ?Sum of angles: (180 - 2θ) + 2φ + angle PSR = 180So, angle PSR = 2θ - 2φBut angle PSR is also part of triangle RSQ. In triangle RSQ, angle at S is 4x, angle at Q is 2θ, and angle at R is 5x.Wait, angle PSR is the same as angle RSQ because both are angles at point S between lines SR and SQ. Wait, no, angle RSQ is at point S between lines RS and SQ, while angle PSR is at point S between lines PS and SR.Wait, maybe they are the same? Because P is on RS, so PS is part of RS. So angle PSR is the same as angle RSQ?Wait, no. Angle RSQ is at point S between RS and SQ, while angle PSR is at point S between PS and SR. Since P is on RS, PS is just a segment of RS, so angle PSR is the same as angle RSQ. Therefore, angle PSR = angle RSQ = 4x degrees.Therefore, from earlier:angle PSR = 2θ - 2φ = 4xSo, 2θ - 2φ = 4xDivide both sides by 2: θ - φ = 2xSo, θ = φ + 2xOkay, so now we have θ expressed in terms of φ.Earlier, we had θ = (180 - 9x)/2So, substituting θ = φ + 2x into that equation:φ + 2x = (180 - 9x)/2Multiply both sides by 2 to eliminate the denominator:2φ + 4x = 180 - 9xBring all terms to one side:2φ + 4x + 9x - 180 = 02φ + 13x - 180 = 02φ = 180 - 13xφ = (180 - 13x)/2Okay, so φ is expressed in terms of x.Now, let's go back to triangle PRS. We had:angle RPS = 180 - 2θangle PRS = 2φangle PSR = 4xWe also know that in triangle PRS, the sum of angles is 180 degrees, so:(180 - 2θ) + 2φ + 4x = 180Simplify:180 - 2θ + 2φ + 4x = 180Subtract 180 from both sides:-2θ + 2φ + 4x = 0Divide both sides by 2:-θ + φ + 2x = 0But we already have θ = φ + 2x, so substituting:-(φ + 2x) + φ + 2x = 0-φ - 2x + φ + 2x = 00 = 0Hmm, that's just a tautology, so it doesn't give us new information. Maybe I need another approach.Let me consider triangle QRP. In triangle QRP, we have:- Angle at R: angle QRP = 5x- Angle at P: angle QPR = 180 - 2θ (from earlier)- Angle at Q: angle RQP = θSum of angles in triangle QRP:5x + (180 - 2θ) + θ = 180Simplify:5x + 180 - 2θ + θ = 1805x + 180 - θ = 180Subtract 180 from both sides:5x - θ = 0So, θ = 5xWait, that's a key equation. So θ equals 5x.But earlier, we had θ = (180 - 9x)/2So, setting them equal:5x = (180 - 9x)/2Multiply both sides by 2:10x = 180 - 9xBring terms together:10x + 9x = 18019x = 180x = 180/19Wait, that's approximately 9.47 degrees, but let me keep it as a fraction.So, x = 180/19 degrees.Now, since θ = 5x, then θ = 5*(180/19) = 900/19 degrees.But angle RQP is θ, so angle RQP = 900/19 degrees.Wait, let me check my steps to make sure I didn't make a mistake.1. In triangle RSQ, angle RSQ = 4x, angle SQR = 2θ, angle QRS = 5x.2. Sum of angles: 4x + 2θ + 5x = 180 => 9x + 2θ = 180 => 2θ = 180 - 9x => θ = (180 - 9x)/2.3. In triangle QRP, angles are 5x (angle QRP), θ (angle RQP), and angle QPR = 180 - 2θ.4. Sum of angles: 5x + θ + (180 - 2θ) = 180 => 5x - θ = 0 => θ = 5x.5. Equate θ expressions: 5x = (180 - 9x)/2 => 10x = 180 - 9x => 19x = 180 => x = 180/19.6. Therefore, θ = 5x = 5*(180/19) = 900/19 ≈ 47.37 degrees.Wait, but earlier I thought angle RQP was θ, which is 5x, but let me confirm.In triangle PQR, since PQ = PR, angles at Q and R are equal, which are angle RQP and angle PRQ. Angle RQP is at point Q, which is θ because QP bisects angle SQR into two θ angles. So yes, angle RQP = θ = 5x.But wait, in triangle QRP, angle RQP is θ, which is 5x, and angle QPR is 180 - 2θ. But angle QPR is also part of triangle PRS, where angle RPS = 180 - 2θ.Wait, but earlier I thought angle PSR = 4x, which is equal to angle RSQ. So that part seems consistent.Let me check if x = 180/19 makes sense.x = 180/19 ≈ 9.47 degrees.Then θ = 5x ≈ 47.37 degrees.Then, in triangle RSQ, angle SQR = 2θ ≈ 94.74 degrees.Angle RSQ = 4x ≈ 37.89 degrees.Angle QRS = 5x ≈ 47.37 degrees.Sum of angles: 94.74 + 37.89 + 47.37 ≈ 180 degrees, which checks out.In triangle QRP, angles are 5x ≈ 47.37, θ ≈ 47.37, and angle QPR = 180 - 2θ ≈ 180 - 94.74 ≈ 85.26 degrees.Sum: 47.37 + 47.37 + 85.26 ≈ 180, which also checks out.In triangle PRS, angle RPS = 180 - 2θ ≈ 85.26 degrees, angle PRS = 2φ, and angle PSR = 4x ≈ 37.89 degrees.Sum: 85.26 + 2φ + 37.89 ≈ 180 => 2φ ≈ 56.85 => φ ≈ 28.42 degrees.From earlier, φ = (180 - 13x)/2.Plugging x = 180/19:φ = (180 - 13*(180/19))/2 = (180 - (2340/19))/2 = (3420/19 - 2340/19)/2 = (1080/19)/2 = 540/19 ≈ 28.42 degrees, which matches.So all the angles check out.Therefore, angle RQP = θ = 5x = 5*(180/19) = 900/19 degrees.Simplifying 900/19, it's approximately 47.37 degrees, but since the question asks for the measure in degrees, we can leave it as a fraction.So, angle RQP = 900/19 degrees.Wait, but let me double-check if angle RQP is indeed θ.In triangle PQR, since PQ = PR, angles at Q and R are equal. Angle at Q is angle RQP, and angle at R is angle PRQ. Both are equal to θ.But angle RQP is also part of the angle bisector QP, which splits angle SQR into two θ angles. So yes, angle RQP = θ.Therefore, the measure of angle RQP is 900/19 degrees, which is approximately 47.37 degrees.But let me express it as a fraction:900 divided by 19 is 47 with a remainder of 7, so 47 and 7/19 degrees, but since the question doesn't specify the form, 900/19 is acceptable.Alternatively, if I made a mistake in assuming angle RQP = θ, maybe I need to reconsider.Wait, angle RQP is at point Q between RQ and PQ. Since QP bisects angle SQR, which is 2θ, then angle PQS = θ and angle PQR = θ. But angle PQR is the same as angle RQP because they are the same angle. So yes, angle RQP = θ.Therefore, my conclusion is correct.**Final Answer**The measure of angle RQP is boxed{dfrac{900}{19}} degrees.

answer：First, I need to determine how many 5-cent and 7-cent stamps Jessica can use to make exactly 50 cents. Let ( s ) represent the number of 5-cent stamps and ( t ) represent the number of 7-cent stamps. The equation to solve is:[5s + 7t = 50]My goal is to minimize the total number of stamps, ( s + t ).I'll start by testing different values of ( t ) to see if they result in a valid integer value for ( s ):1. **When ( t = 0 ):** [ 5s = 50 Rightarrow s = 10 ] Total stamps: ( 10 + 0 = 10 )2. **When ( t = 1 ):** [ 5s = 50 - 7 = 43 ] Since 43 is not a multiple of 5, ( s ) is not an integer.3. **When ( t = 2 ):** [ 5s = 50 - 14 = 36 ] Since 36 is not a multiple of 5, ( s ) is not an integer.4. **When ( t = 3 ):** [ 5s = 50 - 21 = 29 ] Since 29 is not a multiple of 5, ( s ) is not an integer.5. **When ( t = 4 ):** [ 5s = 50 - 28 = 22 ] Since 22 is not a multiple of 5, ( s ) is not an integer.6. **When ( t = 5 ):** [ 5s = 50 - 35 = 15 Rightarrow s = 3 ] Total stamps: ( 3 + 5 = 8 )Among the valid combinations, the minimum number of stamps is 8.

answer：Okay, let's see. I have this function g(x) which is a fifth-degree polynomial: g(x) = dx^5 + ex^3 + fx + 6. I know that when I plug in x = 3, the result is -9. So, g(3) = -9. The question is asking me to find g(-3). Hmm, interesting.First, I should recall what happens when you plug in a negative value into a polynomial. For odd-powered terms like x^5 or x^3, plugging in -x will give me the negative of the original term. For example, (-x)^5 = -x^5 and (-x)^3 = -x^3. On the other hand, the linear term, which is just x, will also become negative when x is negative: f(-x) = -fx. The constant term, which is 6 in this case, remains the same because it doesn't have an x.So, if I write out g(-x), it should be: g(-x) = d(-x)^5 + e(-x)^3 + f(-x) + 6. Simplifying that, it becomes: g(-x) = -dx^5 - ex^3 - fx + 6.Now, if I add g(x) and g(-x) together, let's see what happens. g(x) + g(-x) = (dx^5 + ex^3 + fx + 6) + (-dx^5 - ex^3 - fx + 6). Let's combine like terms:- The dx^5 and -dx^5 cancel each other out.- Similarly, ex^3 and -ex^3 cancel out.- fx and -fx also cancel each other.- The constants 6 and 6 add up to 12.So, g(x) + g(-x) = 12. That's a nice relationship because it tells me that for any x, the sum of g(x) and g(-x) is always 12. Given that g(3) = -9, I can use this relationship to find g(-3). Let's plug x = 3 into the equation:g(3) + g(-3) = 12.We know that g(3) is -9, so substituting that in:-9 + g(-3) = 12.To solve for g(-3), I can add 9 to both sides of the equation:g(-3) = 12 + 9.So, g(-3) = 21.Wait, let me double-check that. If I add -9 and 21, I should get 12. Yes, because -9 + 21 is indeed 12. That makes sense. So, my answer seems correct.I think that's all I need to do here. The key was recognizing the symmetry in the polynomial when evaluating at x and -x, especially since all the exponents are odd, which leads to the cancellation of the variable terms and only the constant term remains. That's pretty cool!**Final Answer**The value of ( g(-3) ) is boxed{21}.

answer：First, I need to determine how much juice each cup receives when the total juice is divided equally among the 5 cups.Given that the container holds (frac{3}{4}) of fruit juice, I divide this amount by 5 to find the quantity in each cup:[frac{3}{4} div 5 = frac{3}{4} times frac{1}{5} = frac{3}{20}]Next, I convert the fraction (frac{3}{20}) into a percentage to find out what percentage of the container's capacity each cup holds:[frac{3}{20} = frac{15}{100} = 15%]Therefore, each cup contains 15% of the container's capacity.