💡First, I need to determine the total number of notebooks available for redistribution. The problem states that there are 1200 notebooks distributed evenly among 30 boxes. To find the total number of notebooks, I can use the initial distribution:[text{Total notebooks} = 1200]Next, the teacher wants to redistribute these notebooks so that each box contains 35 notebooks. To find out how many full boxes can be made with 35 notebooks each, I perform the division:[1200 div 35 = 34 text{ full boxes with a remainder}]To find the exact number of notebooks left over, I calculate the remainder:[1200 mod 35 = 1200 - (35 times 34) = 1200 - 1190 = 10]Therefore, after redistributing the notebooks into as many full boxes of 35 notebooks as possible, there will be 10 notebooks left over.

💡Alright, so I have these two trigonometry problems to solve. Let me take them one at a time. Starting with the first one: Calculate ( sin^2 120^circ + cos 180^circ + tan 45^circ - cos^2 (-330^circ) + sin (-210^circ) ). Hmm, okay. I remember that trigonometric functions have specific values at certain angles, so maybe I can plug those in here. First, let's break down each term individually. 1. ( sin^2 120^circ ): I know that ( 120^circ ) is in the second quadrant, and the sine of 120 degrees is the same as the sine of 60 degrees because it's the reference angle. So, ( sin 60^circ = frac{sqrt{3}}{2} ). Therefore, ( sin^2 120^circ = left( frac{sqrt{3}}{2} right)^2 = frac{3}{4} ).2. ( cos 180^circ ): Cosine of 180 degrees is -1. That's a standard value I remember.3. ( tan 45^circ ): Tangent of 45 degrees is 1. That's straightforward.4. ( cos^2 (-330^circ) ): Cosine is an even function, so ( cos (-330^circ) = cos 330^circ ). Now, 330 degrees is in the fourth quadrant, and its reference angle is 30 degrees. So, ( cos 330^circ = cos 30^circ = frac{sqrt{3}}{2} ). Therefore, ( cos^2 (-330^circ) = left( frac{sqrt{3}}{2} right)^2 = frac{3}{4} ).5. ( sin (-210^circ) ): Sine is an odd function, so ( sin (-210^circ) = -sin 210^circ ). 210 degrees is in the third quadrant, and its reference angle is 30 degrees. In the third quadrant, sine is negative, so ( sin 210^circ = -frac{1}{2} ). Therefore, ( sin (-210^circ) = -(-frac{1}{2}) = frac{1}{2} ).Now, let's substitute all these back into the original expression:( frac{3}{4} + (-1) + 1 - frac{3}{4} + frac{1}{2} ).Let me compute this step by step:- ( frac{3}{4} - 1 = frac{3}{4} - frac{4}{4} = -frac{1}{4} )- ( -frac{1}{4} + 1 = -frac{1}{4} + frac{4}{4} = frac{3}{4} )- ( frac{3}{4} - frac{3}{4} = 0 )- ( 0 + frac{1}{2} = frac{1}{2} )So, the result of the first problem is ( frac{1}{2} ).Moving on to the second problem: Simplify ( frac{sin (alpha - 2pi) cos (alpha - frac{pi}{2}) cos (pi + alpha)}{sin (3pi - alpha) sin (-pi - alpha)} ).Alright, trigonometric identities and angle transformations. Let me recall some properties:- ( sin (theta - 2pi) = sin theta ) because sine has a period of ( 2pi ).- ( cos (alpha - frac{pi}{2}) ): There's an identity for cosine of a difference. ( cos (A - B) = cos A cos B + sin A sin B ). So, ( cos (alpha - frac{pi}{2}) = cos alpha cos frac{pi}{2} + sin alpha sin frac{pi}{2} ). Since ( cos frac{pi}{2} = 0 ) and ( sin frac{pi}{2} = 1 ), this simplifies to ( sin alpha ).- ( cos (pi + alpha) ): Cosine of an angle plus pi is equal to negative cosine of the angle. So, ( cos (pi + alpha) = -cos alpha ).- ( sin (3pi - alpha) ): Let's think about this. ( 3pi - alpha ) is the same as ( pi + (2pi - alpha) ). But maybe it's easier to use the identity ( sin (pi + theta) = -sin theta ). Wait, but here it's ( 3pi - alpha ). Alternatively, ( 3pi - alpha = 2pi + (pi - alpha) ). Since sine has a period of ( 2pi ), ( sin (3pi - alpha) = sin (pi - alpha) ). And ( sin (pi - alpha) = sin alpha ).- ( sin (-pi - alpha) ): Sine is an odd function, so ( sin (-pi - alpha) = -sin (pi + alpha) ). And ( sin (pi + alpha) = -sin alpha ). Therefore, ( sin (-pi - alpha) = -(-sin alpha) = sin alpha ).Now, let's substitute all these back into the expression:Numerator: ( sin alpha cdot sin alpha cdot (-cos alpha) )Denominator: ( sin alpha cdot sin alpha )So, the expression becomes:( frac{sin alpha cdot sin alpha cdot (-cos alpha)}{sin alpha cdot sin alpha} )Simplify numerator and denominator:The ( sin alpha cdot sin alpha ) terms cancel out, leaving:( -cos alpha )So, the simplified form is ( -cos alpha ).**Final Answer**(1) boxed{dfrac{1}{2}}(2) boxed{-cos alpha}

💡Okay, so I have this problem here about an acute-angled triangle with two sides given: one is √13 cm and the other is √10 cm. I need to find the length of the third side, and it's given that this third side is equal to the altitude drawn to it. Hmm, that sounds a bit tricky, but let me try to break it down.First, let me visualize the triangle. Let's call the triangle ABC, where AB is √13 cm, BC is √10 cm, and AC is the third side we need to find. The altitude from B to AC is equal to AC itself. So, if I draw the altitude from point B to side AC, let's call the foot of this altitude D. Then, BD is the altitude, and it's equal to AC. So, BD = AC = x, where x is the length we're trying to find.Now, since BD is the altitude, it splits the triangle into two right-angled triangles: ABD and CBD. Both of these are right-angled at D. So, in triangle ABD, we have AB as the hypotenuse, which is √13 cm, BD as one leg, which is x cm, and AD as the other leg. Similarly, in triangle CBD, BC is the hypotenuse, which is √10 cm, BD is one leg (x cm), and CD is the other leg.Using the Pythagorean theorem on both triangles, I can write two equations:For triangle ABD:AD² + BD² = AB²AD² + x² = (√13)²AD² + x² = 13So, AD² = 13 - x²Therefore, AD = √(13 - x²)For triangle CBD:CD² + BD² = BC²CD² + x² = (√10)²CD² + x² = 10So, CD² = 10 - x²Therefore, CD = √(10 - x²)Now, since AC is the third side, and D is the foot of the altitude from B to AC, AC is equal to AD + CD. So, AC = AD + CD. But AC is also equal to x, so:x = AD + CDx = √(13 - x²) + √(10 - x²)Hmm, okay, so I have this equation: x = √(13 - x²) + √(10 - x²). I need to solve for x. This looks a bit complicated, but maybe I can square both sides to eliminate the square roots.Let me write that down:x = √(13 - x²) + √(10 - x²)Let me denote √(13 - x²) as A and √(10 - x²) as B for simplicity. So, x = A + B.If I square both sides, I get:x² = (A + B)²x² = A² + 2AB + B²But A² is 13 - x² and B² is 10 - x². So, substituting back:x² = (13 - x²) + 2AB + (10 - x²)x² = 13 - x² + 10 - x² + 2ABx² = 23 - 2x² + 2ABNow, let's bring all terms to one side:x² + 2x² - 23 = 2AB3x² - 23 = 2ABBut AB is √(13 - x²) * √(10 - x²). So,3x² - 23 = 2√[(13 - x²)(10 - x²)]Hmm, this still looks complicated, but maybe I can square both sides again to eliminate the square root. Let me try that.First, let me isolate the square root term:(3x² - 23)/2 = √[(13 - x²)(10 - x²)]Now, square both sides:[(3x² - 23)/2]^2 = (13 - x²)(10 - x²)Let me compute both sides.Left side:[(3x² - 23)/2]^2 = (9x⁴ - 138x² + 529)/4Right side:(13 - x²)(10 - x²) = 130 - 13x² - 10x² + x⁴ = x⁴ - 23x² + 130So, now we have:(9x⁴ - 138x² + 529)/4 = x⁴ - 23x² + 130Multiply both sides by 4 to eliminate the denominator:9x⁴ - 138x² + 529 = 4x⁴ - 92x² + 520Bring all terms to the left side:9x⁴ - 138x² + 529 - 4x⁴ + 92x² - 520 = 0Simplify:(9x⁴ - 4x⁴) + (-138x² + 92x²) + (529 - 520) = 05x⁴ - 46x² + 9 = 0Okay, so now I have a quartic equation, but it's quadratic in terms of x². Let me let y = x², so the equation becomes:5y² - 46y + 9 = 0This is a quadratic equation. Let's solve for y using the quadratic formula:y = [46 ± √(46² - 4*5*9)] / (2*5)y = [46 ± √(2116 - 180)] / 10y = [46 ± √1936] / 10√1936 is 44, so:y = [46 ± 44] / 10So, two solutions:y = (46 + 44)/10 = 90/10 = 9y = (46 - 44)/10 = 2/10 = 1/5So, y = 9 or y = 1/5. Since y = x², x² = 9 or x² = 1/5.Therefore, x = √9 = 3 or x = √(1/5) = (√5)/5.But we need to check which of these solutions is valid in the context of the problem.First, let's consider x = 3. Let's check if this makes sense in the original equation:x = √(13 - x²) + √(10 - x²)3 = √(13 - 9) + √(10 - 9)3 = √4 + √13 = 2 + 13 = 3Yes, that works.Now, let's check x = √(1/5). Let's compute:x = √(13 - x²) + √(10 - x²)√(1/5) = √(13 - 1/5) + √(10 - 1/5)√(1/5) = √(64/5) + √(49/5)√(1/5) = (8/√5) + (7/√5)√(1/5) = 15/√5√(1/5) = 3√5But √(1/5) is approximately 0.447, and 3√5 is approximately 6.708. Clearly, 0.447 ≠ 6.708. So, this solution is extraneous and doesn't satisfy the original equation.Therefore, the only valid solution is x = 3 cm.Wait, but the problem mentions it's an acute-angled triangle. Let me make sure that with sides √13, √10, and 3, the triangle is indeed acute.To check if a triangle is acute, we can use the converse of the Pythagorean theorem. For all sides, the square of each side should be less than the sum of the squares of the other two sides.Let's compute:For side 3:3² = 9√10² + √13² = 10 + 13 = 239 < 23, which is true.For side √10:(√10)² = 103² + (√13)² = 9 + 13 = 2210 < 22, which is true.For side √13:(√13)² = 133² + (√10)² = 9 + 10 = 1913 < 19, which is true.Since all sides satisfy the condition, the triangle is indeed acute-angled.So, everything checks out. The length of the third side is 3 cm.**Final Answer**The length of the third side is boxed{3} cm.

💡First, I need to find the complex conjugate of the given complex number (frac{1-3i}{1-i}).To simplify the expression, I'll multiply both the numerator and the denominator by the complex conjugate of the denominator, which is (1+i):[frac{1-3i}{1-i} = frac{(1-3i)(1+i)}{(1-i)(1+i)}]Next, I'll expand both the numerator and the denominator using the distributive property:[frac{(1-3i)(1+i)}{(1-i)(1+i)} = frac{1 cdot 1 + 1 cdot i - 3i cdot 1 - 3i cdot i}{1^2 - i^2}]Simplifying the numerator and the denominator:[frac{1 + i - 3i - 3i^2}{1 - (-1)} = frac{1 - 2i + 3}{2} = frac{4 - 2i}{2} = 2 - i]Now, to find the complex conjugate of (2 - i), I'll change the sign of the imaginary part:[overline{2 - i} = 2 + i]Therefore, the complex conjugate of the given complex number is (2 + i).

💡First, identify the coordinates of point P, which are (4, -3). Here, x = 4 and y = -3.Next, calculate the distance r from the origin to point P using the Pythagorean theorem:r = √(x² + y²) = √(4² + (-3)²) = √(16 + 9) = √25 = 5.Now, use the definition of cosine for any angle:cosα = x / r = 4 / 5.Finally, compare this result with the given options to determine the correct answer.

💡First, I need to simplify the complex number (dfrac{1+2i}{i}). To do this, I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is (-i), to eliminate the imaginary unit from the denominator.[dfrac{1+2i}{i} = dfrac{(1+2i)(-i)}{i cdot (-i)} = dfrac{-i - 2i^2}{-i^2}]Since (i^2 = -1), this simplifies to:[dfrac{-i - 2(-1)}{-(-1)} = dfrac{-i + 2}{1} = 2 - i]Now that I have the simplified form (2 - i), I'll find its conjugate by changing the sign of the imaginary part:[text{Conjugate of } 2 - i = 2 + i]This means (a = 2) and (b = 1). Therefore, the point ((a, b)) is ((2, 1)).