💡First, I need to find the slope of the line from the given equation (2x + ay + 3 = 0). I'll rearrange the equation to the slope-intercept form (y = mx + b).Starting with:[2x + ay + 3 = 0]Subtract (2x) and (3) from both sides:[ay = -2x - 3]Divide both sides by (a):[y = -frac{2}{a}x - frac{3}{a}]Now, the slope (m) of the line is (-frac{2}{a}).Next, I know that the slope angle (theta) is related to the slope by the equation:[m = tan(theta)]Given that the slope angle is (120^circ), I can find the tangent of (120^circ):[tan(120^circ) = tan(180^circ - 60^circ) = -tan(60^circ) = -sqrt{3}]So, the slope (m) is:[m = -sqrt{3}]Setting the two expressions for (m) equal to each other:[-frac{2}{a} = -sqrt{3}]Simplify by multiplying both sides by (-1):[frac{2}{a} = sqrt{3}]Solve for (a):[a = frac{2}{sqrt{3}}]To rationalize the denominator, multiply numerator and denominator by (sqrt{3}):[a = frac{2sqrt{3}}{3}]

💡Okay, so I have this problem where I need to calculate the angle θ in degrees for the expression:cis 60° + cis 70° + cis 80° + … + cis 140°and express it in the form r cis θ, where r > 0 and 0° ≤ θ < 360°. Hmm, okay. I remember that cis θ is shorthand for cos θ + i sin θ, right? So, each term in the sum is a complex number on the unit circle at the corresponding angle.First, let me figure out how many terms are in this sum. It starts at 60° and goes up by 10° each time until it reaches 140°. So, from 60° to 140°, that's 60, 70, 80, 90, 100, 110, 120, 130, 140. Let me count: 60 is the first, then 70, 80, 90, 100, 110, 120, 130, 140. That's 9 terms in total.Now, I need to add all these complex numbers together. Adding complex numbers can be done by adding their real parts and their imaginary parts separately. So, if I write each term as cos θ + i sin θ, then the sum will be the sum of all the cosines plus i times the sum of all the sines.So, let me write that out:Sum = (cos 60° + cos 70° + cos 80° + … + cos 140°) + i (sin 60° + sin 70° + sin 80° + … + sin 140°)Now, I need to compute both the real part (sum of cosines) and the imaginary part (sum of sines). This might be a bit tedious, but maybe there's a pattern or a formula I can use to simplify this.I recall that the sum of a series of complex numbers in the form cis θ can be expressed using the formula for the sum of a geometric series if the angles form an arithmetic sequence. In this case, the angles do form an arithmetic sequence with a common difference of 10°, starting at 60° and ending at 140°, which we already established has 9 terms.The formula for the sum of a geometric series is S = a₁ (1 - rⁿ) / (1 - r), where a₁ is the first term, r is the common ratio, and n is the number of terms. In this case, each term is cis θ, so the common ratio r would be cis 10°, since each term is 10° more than the previous one.So, applying the formula, the sum S would be:S = cis 60° (1 - (cis 10°)⁹) / (1 - cis 10°)Hmm, okay. Let me compute (cis 10°)⁹ first. Using De Moivre's theorem, (cis θ)ⁿ = cis (nθ). So, (cis 10°)⁹ = cis 90°, which is cos 90° + i sin 90° = 0 + i(1) = i.So, substituting back into the formula:S = cis 60° (1 - i) / (1 - cis 10°)Now, let me compute the denominator: 1 - cis 10°. Using cis θ = cos θ + i sin θ, so 1 - cis 10° = 1 - cos 10° - i sin 10°.So, the denominator is 1 - cos 10° - i sin 10°, and the numerator is cis 60° (1 - i). Let me compute cis 60° first. That's cos 60° + i sin 60° = 0.5 + i (√3/2).So, multiplying cis 60° by (1 - i):(0.5 + i √3/2)(1 - i) = 0.5(1) + 0.5(-i) + i √3/2 (1) + i √3/2 (-i)Simplify each term:= 0.5 - 0.5i + (√3/2)i - (√3/2)i²Remember that i² = -1, so:= 0.5 - 0.5i + (√3/2)i + (√3/2)Combine like terms:Real parts: 0.5 + √3/2Imaginary parts: (-0.5 + √3/2)iSo, the numerator becomes:(0.5 + √3/2) + i (-0.5 + √3/2)Now, the denominator is 1 - cos 10° - i sin 10°. Let me compute 1 - cos 10° and sin 10° numerically to make it easier.First, cos 10° ≈ 0.9848, so 1 - cos 10° ≈ 1 - 0.9848 = 0.0152Sin 10° ≈ 0.1736So, the denominator is approximately 0.0152 - i 0.1736Now, to divide the numerator by the denominator, I can multiply numerator and denominator by the complex conjugate of the denominator to rationalize it.The complex conjugate of the denominator is 0.0152 + i 0.1736So, multiplying numerator and denominator by this:Numerator: [(0.5 + √3/2) + i (-0.5 + √3/2)] * (0.0152 + i 0.1736)Denominator: (0.0152 - i 0.1736)(0.0152 + i 0.1736) = (0.0152)² + (0.1736)² ≈ 0.000231 + 0.03014 ≈ 0.03037Now, let's compute the numerator:Let me denote A = 0.5 + √3/2 ≈ 0.5 + 0.8660 ≈ 1.3660B = -0.5 + √3/2 ≈ -0.5 + 0.8660 ≈ 0.3660So, the numerator becomes:(A + i B)(0.0152 + i 0.1736) = A*0.0152 + A*i 0.1736 + i B*0.0152 + i² B*0.1736= A*0.0152 + i A*0.1736 + i B*0.0152 - B*0.1736Combine real and imaginary parts:Real: A*0.0152 - B*0.1736Imaginary: (A*0.1736 + B*0.0152)iCompute each:A ≈ 1.3660, B ≈ 0.3660Real: 1.3660*0.0152 ≈ 0.02076 - 0.3660*0.1736 ≈ 0.0636Wait, that doesn't seem right. Let me recalculate:Real part: A*0.0152 - B*0.1736 ≈ 1.3660*0.0152 ≈ 0.02076 - 0.3660*0.1736 ≈ 0.0636Wait, 0.02076 - 0.0636 ≈ -0.04284Imaginary part: A*0.1736 + B*0.0152 ≈ 1.3660*0.1736 ≈ 0.2373 + 0.3660*0.0152 ≈ 0.00556 ≈ 0.2429So, the numerator is approximately (-0.04284) + i 0.2429Denominator is approximately 0.03037So, dividing numerator by denominator:(-0.04284 / 0.03037) + i (0.2429 / 0.03037) ≈ (-1.410) + i (7.995)So, approximately, the sum S ≈ -1.410 + i 8.0Wait, that seems quite large. Let me check my calculations because I might have made a mistake somewhere.Wait, when I computed the numerator after multiplying by the conjugate, I think I might have messed up the signs or the multiplication.Let me try a different approach. Maybe instead of using the geometric series formula, I can use the formula for the sum of sines and cosines in an arithmetic sequence.I remember that the sum of cos(a) + cos(a + d) + ... + cos(a + (n-1)d) can be expressed as [sin(n d / 2) / sin(d / 2)] * cos(a + (n - 1)d / 2)Similarly, the sum of sin(a) + sin(a + d) + ... + sin(a + (n-1)d) is [sin(n d / 2) / sin(d / 2)] * sin(a + (n - 1)d / 2)So, in this case, a = 60°, d = 10°, n = 9.So, let's compute the sum of cosines:Sum_cos = [sin(9 * 10° / 2) / sin(10° / 2)] * cos(60° + (9 - 1)*10° / 2)Simplify:Sum_cos = [sin(45°) / sin(5°)] * cos(60° + 40°) = [sin(45°)/sin(5°)] * cos(100°)Similarly, Sum_sin = [sin(45°)/sin(5°)] * sin(100°)So, both sums have the same factor [sin(45°)/sin(5°)] multiplied by cos(100°) and sin(100°) respectively.So, the total sum S = Sum_cos + i Sum_sin = [sin(45°)/sin(5°)] * (cos(100°) + i sin(100°)) = [sin(45°)/sin(5°)] cis(100°)Therefore, the angle θ is 100°, and the magnitude r is [sin(45°)/sin(5°)].Wait, that seems more straightforward. So, the angle θ is 100°, which is between 0° and 360°, so that's our answer.I think my initial approach with the geometric series was correct, but maybe I made a calculation error when trying to compute it numerically. The second method using the sum formulas for sine and cosine in arithmetic sequences seems more reliable and gives θ = 100°, which makes sense because the angles are symmetrically distributed around 100°, so their vector sum should point in that direction.

💡Okay, so I have this math problem here: 5 times (9 minus 3) plus 8. Hmm, let me try to figure this out step by step. First, I remember that when solving expressions, I should follow the order of operations, which is often remembered by the acronym PEMDAS—Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction.Alright, looking at the expression, there are parentheses here: (9 - 3). So according to PEMDAS, I should start by solving what's inside the parentheses. Let me subtract 3 from 9. That gives me 6. So now, the expression simplifies to 5 times 6 plus 8.Next, I need to handle the multiplication. So, 5 multiplied by 6. Let me do that: 5 times 6 is 30. Now, the expression is down to 30 plus 8.Finally, I just need to add 8 to 30. That should be straightforward. 30 plus 8 equals 38. So, putting it all together, the value of the expression 5 times (9 minus 3) plus 8 is 38.Wait, let me double-check to make sure I didn't make any mistakes. Starting again: inside the parentheses, 9 minus 3 is definitely 6. Then, 5 times 6 is 30. Adding 8 to 30 gives me 38. Yep, that seems right. I think I've got it.

💡First, I'll assume the fuel tank has a capacity of 1 unit to simplify calculations.1. **Initial Fill with Brand Z Gasoline:** - The tank is completely filled with brand Z gasoline, so it contains 1 unit of brand Z.2. **First Emptying and Refilling with Brand X Gasoline:** - The tank is emptied to 3/4 of its capacity, leaving 1/4 unit of brand Z. - It is then filled with brand X gasoline, resulting in 1/4 unit of brand Z and 3/4 unit of brand X.3. **Second Emptying and Refilling with Brand Z Gasoline:** - The tank is emptied to 2/3 of its capacity, leaving 1/3 of the remaining brand Z, which is 1/12 unit. - It is then filled with brand Z gasoline, bringing the total brand Z back to 1 unit.4. **Third Emptying and Refilling with Brand X Gasoline:** - The tank is emptied to 5/8 of its capacity, leaving 3/8 unit of brand Z. - It is then filled with brand X gasoline, resulting in 3/8 unit of brand Z and 5/8 unit of brand X.5. **Fourth Emptying and Refilling with Brand Z Gasoline:** - The tank is emptied to 4/5 of its capacity, leaving 1/5 of the remaining brand Z, which is 3/40 unit. - It is then filled with brand Z gasoline, bringing the total brand Z back to 1 unit.6. **Fifth Emptying and Refilling with Brand X Gasoline:** - The tank is emptied to 7/8 of its capacity, leaving 1/8 unit of brand Z. - It is then filled with brand X gasoline, resulting in 1/8 unit of brand Z and 7/8 unit of brand X.Finally, to determine the percentage of brand Z gasoline in the tank:- The tank has 1/8 unit of brand Z out of a total of 1 unit.- This is equivalent to 12.5% of the gasoline in the tank being brand Z.

💡Alright, so I've got this vector problem here. Let me try to wrap my head around it step by step. The problem says:"Given that point ( M ) lies in the plane ( ABC ), and for any point ( O ) in space, ( overrightarrow{OM} = xoverrightarrow{OA} + frac{1}{3}overrightarrow{OB} + frac{1}{2}overrightarrow{OC} ), then ( x = ) ( )"And the options are:A: ( -frac{1}{6} )B: ( frac{1}{6} )C: ( frac{1}{2} )D: ( frac{1}{3} )Okay, so first, let's parse what this means. We have point ( M ) lying in the plane formed by points ( A ), ( B ), and ( C ). For any point ( O ) in space, the vector from ( O ) to ( M ) is given as a linear combination of vectors from ( O ) to ( A ), ( B ), and ( C ). The coefficients of these vectors are ( x ), ( frac{1}{3} ), and ( frac{1}{2} ) respectively.I remember that when a point lies in the plane formed by three other points, the position vector of that point can be expressed as a linear combination of the position vectors of the three points, with the coefficients adding up to 1. This is because the point is a barycentric combination of the three points.So, in this case, since ( M ) is in the plane ( ABC ), the coefficients ( x ), ( frac{1}{3} ), and ( frac{1}{2} ) should add up to 1. That seems like a key insight here.Let me write that down:( x + frac{1}{3} + frac{1}{2} = 1 )Now, I need to solve for ( x ). To do that, I'll first find a common denominator for the fractions to make the addition easier. The denominators here are 3 and 2, so the least common denominator is 6.Converting each fraction:( frac{1}{3} = frac{2}{6} )( frac{1}{2} = frac{3}{6} )So, substituting back into the equation:( x + frac{2}{6} + frac{3}{6} = 1 )Now, combining the fractions:( x + frac{5}{6} = 1 )To isolate ( x ), I'll subtract ( frac{5}{6} ) from both sides:( x = 1 - frac{5}{6} )Calculating the right side:( 1 = frac{6}{6} ), so( x = frac{6}{6} - frac{5}{6} = frac{1}{6} )So, ( x = frac{1}{6} ). Looking back at the options, that's option B.Wait, let me double-check to make sure I didn't make a mistake. Sometimes with fractions, it's easy to slip up.Starting again:( x + frac{1}{3} + frac{1}{2} = 1 )Convert to sixths:( x + frac{2}{6} + frac{3}{6} = 1 )Combine the fractions:( x + frac{5}{6} = 1 )Subtract ( frac{5}{6} ):( x = frac{1}{6} )Yep, that seems consistent. So, I think I did it right.Another way to think about it is that since ( M ) is in the plane ( ABC ), the coefficients must satisfy the condition that their sum is 1. This ensures that ( M ) is a weighted average of points ( A ), ( B ), and ( C ), keeping it within the plane.So, adding up ( x ), ( frac{1}{3} ), and ( frac{1}{2} ) must equal 1, which gives ( x = frac{1}{6} ).I don't see any other considerations here because the problem states that this holds for any point ( O ) in space. That means the relationship is independent of the choice of ( O ), which reinforces that the coefficients must satisfy the condition of summing to 1.Therefore, I'm confident that the correct answer is ( frac{1}{6} ), which is option B.

💡Alright, so I've got this inequality system to solve for x:[begin{cases}x + 5 > 0 x - m leq 1end{cases}]And I need to find the range of the real number m such that this system has exactly 3 integer solutions. Hmm, okay. Let me break this down step by step.First, I'll tackle each inequality separately to understand what they're telling me about x.Starting with the first inequality: (x + 5 > 0). If I subtract 5 from both sides, I get (x > -5). So, x has to be greater than -5. That means x can be -4, -3, -2, -1, 0, and so on. But since we're dealing with integer solutions, I'm thinking of integers greater than -5.Now, moving on to the second inequality: (x - m leq 1). If I add m to both sides, it becomes (x leq m + 1). So, x has to be less than or equal to (m + 1). That means x can be at most (m + 1). Depending on the value of m, this upper bound will change.So, combining both inequalities, x has to satisfy both (x > -5) and (x leq m + 1). Therefore, the solution set for x is the interval ((-5, m + 1]). Now, since we're looking for integer solutions, I need to find the integers that lie within this interval.The problem states that there are exactly 3 integer solutions. So, I need to figure out what values of m will make sure that there are exactly 3 integers between -5 and (m + 1).Let me think about what integers could be in this interval. If x has to be greater than -5, the smallest integer x can be is -4. Then, the next integers are -3, -2, -1, 0, and so on. Since we need exactly 3 integer solutions, I need to identify which three integers these could be.If I consider the smallest possible integers, they would be -4, -3, and -2. That's three integers. So, if the upper bound (m + 1) is just enough to include -2 but not -1, that should give me exactly 3 integer solutions.So, if (m + 1) is greater than or equal to -2 but less than -1, then x can be -4, -3, and -2, but not -1. That would mean:[-2 leq m + 1 < -1]To solve for m, I'll subtract 1 from all parts of the inequality:[-2 - 1 leq m < -1 - 1][-3 leq m < -2]So, m has to be greater than or equal to -3 and less than -2.Wait, let me double-check that. If m is -3, then (m + 1 = -2). So, the inequality becomes (x leq -2). Combining with (x > -5), the integer solutions are -4, -3, -2. That's three integers.If m is just a little less than -2, say -2.1, then (m + 1 = -1.1). So, (x leq -1.1). But since x has to be an integer, the largest integer less than or equal to -1.1 is -2. So, again, the integer solutions are -4, -3, -2. That's still three integers.But if m is exactly -2, then (m + 1 = -1). So, (x leq -1). Then, the integer solutions would be -4, -3, -2, -1. That's four integers, which is more than we want. So, m can't be -2 or less than -2? Wait, no, if m is less than -2, like -2.5, then (m + 1 = -1.5), so x has to be less than or equal to -1.5, which still gives us integer solutions up to -2. So, m has to be greater than or equal to -3 and less than -2 to ensure that the integer solutions are exactly -4, -3, -2.Let me visualize this on a number line. The interval for x is from just above -5 to (m + 1). If (m + 1) is between -2 and -1, then the integers in that interval are -4, -3, -2. If (m + 1) is equal to -2, then -2 is included, but -1 is not. If (m + 1) is less than -2, say -3, then the upper bound is lower, but since m is a real number, (m + 1) can be any real number between -5 and infinity, but we need it to be just right to include exactly three integers.Wait, actually, if (m + 1) is less than -2, say -2.5, then the upper bound is -2.5, so the integers less than or equal to -2.5 are -3, -4, etc., but since x has to be greater than -5, the integers would be -4, -3, -2. So, even if (m + 1) is less than -2, as long as it's greater than or equal to -3, we still get three integer solutions.Wait, no. If (m + 1) is less than -2, say -2.5, then (x leq -2.5). So, the integers less than or equal to -2.5 are -3, -4, -5, etc., but x has to be greater than -5, so the integers are -4, -3, -2. Wait, no, -2 is greater than -2.5, so -2 is not included. So, if (m + 1) is -2.5, then x can be -4, -3, -2.5, but since x has to be an integer, the integer solutions are -4, -3. That's only two integers, which is less than we need.Wait, I'm getting confused. Let me clarify.If (m + 1) is -2.5, then (x leq -2.5). So, the integers less than or equal to -2.5 are -3, -4, -5, etc. But x has to be greater than -5, so the integers are -4, -3. That's only two integers. So, to have three integer solutions, (m + 1) has to be at least -2, so that -2 is included. But if (m + 1) is exactly -2, then x can be -4, -3, -2. If (m + 1) is greater than -2, say -1.5, then x can be -4, -3, -2, -1, which is four integers. So, to have exactly three integer solutions, (m + 1) has to be between -2 and -1, but not including -1. So, ( -2 leq m + 1 < -1 ).Therefore, solving for m:[-2 leq m + 1 < -1]Subtract 1 from all parts:[-3 leq m < -2]So, m has to be greater than or equal to -3 and less than -2.Let me test this with m = -3. Then, (m + 1 = -2). So, x has to be greater than -5 and less than or equal to -2. The integers are -4, -3, -2. That's three integers.If m = -2.5, then (m + 1 = -1.5). So, x has to be greater than -5 and less than or equal to -1.5. The integers are -4, -3, -2. That's still three integers.If m approaches -2 from below, say m = -2.1, then (m + 1 = -1.1). So, x has to be greater than -5 and less than or equal to -1.1. The integers are -4, -3, -2. Still three integers.If m = -2, then (m + 1 = -1). So, x has to be greater than -5 and less than or equal to -1. The integers are -4, -3, -2, -1. That's four integers, which is too many.If m is less than -3, say m = -4, then (m + 1 = -3). So, x has to be greater than -5 and less than or equal to -3. The integers are -4, -3. That's only two integers, which is too few.Therefore, the range of m that gives exactly three integer solutions is from -3 up to, but not including, -2.So, the range of m is (-3 leq m < -2).**Final Answer**The range of real number ( m ) is boxed{-3 leq m < -2}.