💡To solve the problem of scheduling 4 mathematics courses in a 7-period day without any two courses being consecutive, we can use the principle of inclusion-exclusion.First, we calculate the total number of ways to schedule the 4 courses without any restrictions. This is done by choosing 4 periods out of 7 and then arranging the 4 courses in those periods:[binom{7}{4} times 4! = 35 times 24 = 840 text{ ways}]Next, we subtract the cases where at least two courses are consecutive. We consider two cases: at least two consecutive courses and all three consecutive courses.**Case 1: At least two consecutive courses**We treat two consecutive periods as a "block." There are 6 possible positions for this block. After placing this block, we have 5 remaining periods to place the third and fourth course. The two courses in the block can be arranged in:[binom{4}{2} times 2! = 6 times 2 = 12 text{ ways}]The third and fourth courses can be placed in any of the 5 remaining periods in:[binom{5}{2} times 2! = 10 times 2 = 20 text{ ways}]Thus, the total number of ways for this case is:[6 times 12 times 20 = 1440 text{ ways}]**Case 2: All three consecutive courses**We treat three consecutive periods as a "block." There are 5 possible positions for this block. The three courses can be arranged in this block in:[binom{4}{3} times 3! = 4 times 6 = 24 text{ ways}]The fourth course can be placed in any of the 4 remaining periods. Thus, the total number of ways for this case is:[5 times 24 times 4 = 480 text{ ways}]Using the principle of inclusion-exclusion, we subtract these invalid cases from the total number of ways:[840 - 1440 + 480 = -120 + 480 = 360]Thus, the number of valid ways to schedule the 4 mathematics courses is:[boxed{textbf{(A)} , 360}]

💡First, I'll define the number of marbles each person has:- Atticus: A- Jensen: J- Cruz: C- Harper: HFrom the problem, I know that:1. Four times the sum of their marbles equals 120: 4 * (A + J + C + H) = 1202. Atticus has half as many marbles as Jensen: A = 1/2 * J3. Atticus has 4 marbles: A = 44. Jensen has twice as many marbles as Harper: J = 2 * HUsing the third point, since A = 4, and from the second point, A = 1/2 * J, I can find J:4 = 1/2 * JJ = 8Now, using the fourth point, J = 2 * H, and knowing J = 8:8 = 2 * HH = 4Now, I'll substitute the known values of A, J, and H into the first equation to find C:4 * (4 + 8 + C + 4) = 1204 * (16 + C) = 12016 + C = 30C = 14Therefore, Cruz has 14 marbles.

💡Okay, so I have to simplify sin 45 degrees minus sin 15 degrees. Hmm, I remember there are some trigonometric identities that can help with this. Let me think... I think the sine of a difference formula might be useful here, or maybe the sum-to-product identities. Wait, the problem is asking for the difference of two sines, so maybe the sum-to-product identity is the way to go.Let me recall the sum-to-product identities. I think one of them is sin A minus sin B equals something. Let me see... I believe it's 2 times cos of (A plus B)/2 times sin of (A minus B)/2. Yeah, that sounds right. So, if I let A be 45 degrees and B be 15 degrees, then I can apply this identity.So, plugging in the values, sin 45° - sin 15° should be equal to 2 times cos of (45° + 15°)/2 times sin of (45° - 15°)/2. Let me compute those angles inside the cosine and sine functions.First, (45° + 15°)/2 is (60°)/2, which is 30°. Then, (45° - 15°)/2 is (30°)/2, which is 15°. So, now the expression becomes 2 times cos 30° times sin 15°. Okay, that seems simpler, but I still need to compute cos 30° and sin 15°.I remember that cos 30° is one of those standard angles. Let me recall, cos 30° is √3/2. Right, that's correct. Now, sin 15°, that's a bit trickier. I don't remember the exact value off the top of my head, but I think I can use another identity to find it. Maybe the sine of a difference formula?Yes, sin(A - B) equals sin A cos B minus cos A sin B. So, if I let A be 45° and B be 30°, then sin 15° is sin(45° - 30°). So, applying the formula, sin 45° cos 30° minus cos 45° sin 30°. I know sin 45° is √2/2, cos 30° is √3/2, cos 45° is √2/2, and sin 30° is 1/2.So, plugging those in, sin 15° is (√2/2)(√3/2) minus (√2/2)(1/2). Let me compute each term. The first term is (√2 * √3)/(2 * 2) which is √6/4. The second term is (√2 * 1)/(2 * 2) which is √2/4. So, sin 15° is √6/4 minus √2/4, which can be written as (√6 - √2)/4.Alright, so going back to the earlier expression, sin 45° - sin 15° is equal to 2 times cos 30° times sin 15°. We have cos 30° as √3/2 and sin 15° as (√6 - √2)/4. So, plugging those in, it becomes 2 times (√3/2) times (√6 - √2)/4.Let me compute this step by step. First, 2 times (√3/2) is just √3. Then, √3 times (√6 - √2)/4. So, that's (√3 * √6 - √3 * √2)/4. Let me simplify those radicals.√3 * √6 is √(3*6) which is √18. √18 can be simplified to 3√2 because 18 is 9*2 and √9 is 3. Similarly, √3 * √2 is √6. So, putting it all together, we have (3√2 - √6)/4.Wait a second, let me double-check that. So, √3 * √6 is √18 which is 3√2, and √3 * √2 is √6. So, yes, that part seems correct. So, the numerator is 3√2 - √6, and the denominator is 4. Therefore, the entire expression simplifies to (3√2 - √6)/4.Let me just verify this result another way to make sure I didn't make a mistake. Alternatively, I could compute sin 45° and sin 15° numerically and subtract them, then see if it matches (3√2 - √6)/4.So, sin 45° is approximately 0.7071, and sin 15° is approximately 0.2588. Subtracting them gives roughly 0.7071 - 0.2588 = 0.4483.Now, let's compute (3√2 - √6)/4 numerically. √2 is approximately 1.4142, so 3√2 is about 4.2426. √6 is approximately 2.4495. So, 4.2426 - 2.4495 is approximately 1.7931. Dividing that by 4 gives approximately 0.4483, which matches the earlier result. So, that seems to confirm that my simplification is correct.Therefore, I can confidently say that sin 45° - sin 15° simplifies to (3√2 - √6)/4.

💡First, I need to determine how many tomatoes Uncle Jerry reaped today. I know that he reaped a total of 290 tomatoes over the two days and that he reaped 120 tomatoes yesterday.To find out how many tomatoes he reaped today, I subtract the number of tomatoes reaped yesterday from the total number of tomatoes:290 - 120 = 170 tomatoes.Now, to find out how many more tomatoes he reaped today compared to yesterday, I subtract the number of tomatoes reaped yesterday from the number reaped today:170 - 120 = 50 more tomatoes.Therefore, Uncle Jerry reaped 50 more tomatoes today compared to yesterday.

💡Okay, so I've got these two equations to solve. Let's take them one at a time. The first one is a quadratic equation: ( x^2 - 5x - 6 = 0 ). I remember that quadratic equations can often be solved by factoring, completing the square, or using the quadratic formula. I think factoring might work here because the numbers seem manageable.Alright, to factor ( x^2 - 5x - 6 ), I need two numbers that multiply to -6 and add up to -5. Hmm, let's see. The factors of -6 could be -6 and 1, or 6 and -1. Let me check: -6 and 1 multiply to -6 and add up to -5. Perfect! So, the equation factors to ( (x - 6)(x + 1) = 0 ).Setting each factor equal to zero gives the solutions:1. ( x - 6 = 0 ) leads to ( x = 6 ).2. ( x + 1 = 0 ) leads to ( x = -1 ).So, the solutions for the first equation are ( x = 6 ) and ( x = -1 ).Now, moving on to the second equation: ( (x + 1)(x - 1) + x(x + 2) = 7 + 6x ). This looks a bit more complicated, but I think I can simplify it step by step.First, I'll expand the left side:- ( (x + 1)(x - 1) ) is a difference of squares, so it becomes ( x^2 - 1 ).- ( x(x + 2) ) expands to ( x^2 + 2x ).Adding these together, the left side becomes ( x^2 - 1 + x^2 + 2x ), which simplifies to ( 2x^2 + 2x - 1 ).Now, the equation is ( 2x^2 + 2x - 1 = 7 + 6x ). I'll move all terms to one side to set the equation to zero:- Subtract ( 7 + 6x ) from both sides: ( 2x^2 + 2x - 1 - 7 - 6x = 0 ).- Simplify: ( 2x^2 - 4x - 8 = 0 ).To make it simpler, I'll divide the entire equation by 2:- ( x^2 - 2x - 4 = 0 ).Now, this is a quadratic equation that doesn't factor nicely, so I'll use the quadratic formula: ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -2 ), and ( c = -4 ).Plugging in the values:- Discriminant: ( (-2)^2 - 4(1)(-4) = 4 + 16 = 20 ).- So, ( x = frac{2 pm sqrt{20}}{2} ).- Simplify ( sqrt{20} ) to ( 2sqrt{5} ), so ( x = frac{2 pm 2sqrt{5}}{2} ).- Divide numerator and denominator by 2: ( x = 1 pm sqrt{5} ).Therefore, the solutions are ( x = 1 + sqrt{5} ) and ( x = 1 - sqrt{5} ).Wait, let me double-check my steps to make sure I didn't make any mistakes. For the first equation, factoring seemed straightforward, and the solutions make sense. For the second equation, expanding and simplifying led me to a quadratic that required the quadratic formula. The discriminant was positive, so two real solutions, which matches my results. I think everything looks correct.

💡Alright, so I've got this geometry problem here. Let me try to visualize it first. There's a larger square, and inside it, there's a smaller square right in the center. Around this smaller square, there are four identical rectangles placed. So, the larger square is made up of the smaller square plus these four rectangles.The problem says that the area of the larger, outer square is four times that of the smaller, inner square. I need to find the ratio of the length of the longer side to the shorter side of each rectangle. The options are given as A) 1.5, B) 2, C) 2.5, D) 3, and E) 3.5.Okay, let's start by assigning some variables to make this concrete. Let me denote the side length of the smaller, inner square as 's'. Since the area of the larger square is four times that of the smaller one, the area of the larger square must be 4 times s squared, which is 4s². Therefore, the side length of the larger square must be the square root of 4s², which is 2s. So, the larger square has a side length of 2s.Now, the four identical rectangles are placed around the smaller square. Let me think about how they are arranged. If I imagine the smaller square in the center, each side of the smaller square must be adjacent to one of these rectangles. So, each rectangle is attached to one side of the smaller square.Let me denote the shorter side of each rectangle as 'y' and the longer side as 'x'. Since the rectangles are identical and placed symmetrically around the smaller square, the dimensions should fit perfectly to form the larger square.Looking at the larger square, its side length is 2s. This side length is composed of the side of the smaller square plus two times the shorter side of the rectangle. Wait, is that correct? Let me think.If I look at the larger square, on each side, it's made up of the smaller square's side and two rectangles attached to it. But actually, since the rectangles are placed on all four sides, perhaps the side length of the larger square is the sum of the smaller square's side and two times the shorter side of the rectangle. Hmm, that seems plausible.So, mathematically, that would be:s + 2y = 2sWait, that simplifies to 2y = s, so y = s/2. Okay, so the shorter side of each rectangle is half the side length of the smaller square.Now, what about the longer side of the rectangle? Since the rectangles are placed adjacent to the smaller square, the longer side must align with the side of the larger square. So, the longer side of the rectangle plus the side of the smaller square should equal the side length of the larger square.So, that would be:x + s = 2sSolving for x, we get x = s. So, the longer side of each rectangle is equal to the side length of the smaller square.Now, the ratio of the longer side to the shorter side of each rectangle is x/y. Substituting the values we found, that's s divided by (s/2), which simplifies to 2.So, the ratio is 2, which corresponds to option B.Wait, let me double-check to make sure I didn't make any mistakes. The area of the larger square is four times the smaller one, so the side length is double. The shorter side of the rectangle is half the smaller square's side, and the longer side is equal to the smaller square's side. Therefore, the ratio is indeed 2. Yeah, that seems right.