💡Alright, so I've got this equation here: ( x^2 + 3y^2 - 4x - 12y + 28 = 0 ). I need to figure out what kind of graph this represents. The options are parabola, circle, ellipse, hyperbola, point, line, two lines, or empty. Hmm, okay.First, I remember that the general form of a conic section is ( Ax^2 + By^2 + Cx + Dy + E = 0 ). Depending on the coefficients and how they relate to each other, it can represent different conic sections. So, let's see.Looking at the equation, I see both ( x^2 ) and ( y^2 ) terms, which suggests it's not a parabola because parabolas only have one squared term. It could be a circle, ellipse, hyperbola, or maybe something else.Since both ( x^2 ) and ( y^2 ) are present and have positive coefficients, it might be a circle or an ellipse. But the coefficients are different: 1 for ( x^2 ) and 3 for ( y^2 ). That makes me think it's not a circle because in a circle, the coefficients of ( x^2 ) and ( y^2 ) are equal. So, maybe it's an ellipse.But before jumping to conclusions, I should try completing the square for both ( x ) and ( y ) to see if I can rewrite the equation in a standard form.Starting with the ( x )-terms: ( x^2 - 4x ). To complete the square, I take half of the coefficient of ( x ), which is -4, so half of that is -2, and then square it, getting 4. So, ( x^2 - 4x ) becomes ( (x - 2)^2 - 4 ).Now for the ( y )-terms: ( 3y^2 - 12y ). I notice there's a coefficient in front of ( y^2 ), so I should factor that out first. Factoring out 3, I get ( 3(y^2 - 4y) ). Now, completing the square inside the parentheses: take half of -4, which is -2, square it to get 4. So, ( y^2 - 4y ) becomes ( (y - 2)^2 - 4 ). Therefore, the entire ( y )-terms become ( 3[(y - 2)^2 - 4] = 3(y - 2)^2 - 12 ).Now, substituting these back into the original equation:( (x - 2)^2 - 4 + 3(y - 2)^2 - 12 + 28 = 0 )Let's simplify this:( (x - 2)^2 + 3(y - 2)^2 - 4 - 12 + 28 = 0 )Combine the constants:-4 -12 = -16-16 + 28 = 12So, the equation becomes:( (x - 2)^2 + 3(y - 2)^2 + 12 = 0 )Wait, that doesn't look right. If I move the 12 to the other side:( (x - 2)^2 + 3(y - 2)^2 = -12 )Hmm, the left side is a sum of squares, which are always non-negative, and the right side is negative. That can't happen because the sum of non-negative numbers can't be negative. So, does that mean there are no real solutions? That would imply the graph is empty.But let me double-check my steps to make sure I didn't make a mistake.Starting again:Original equation: ( x^2 + 3y^2 - 4x - 12y + 28 = 0 )Group ( x ) and ( y ) terms:( x^2 - 4x + 3y^2 - 12y + 28 = 0 )Complete the square for ( x ):( x^2 - 4x = (x - 2)^2 - 4 )Complete the square for ( y ):Factor out 3: ( 3(y^2 - 4y) )Complete the square inside: ( y^2 - 4y = (y - 2)^2 - 4 )So, ( 3(y^2 - 4y) = 3[(y - 2)^2 - 4] = 3(y - 2)^2 - 12 )Substitute back:( (x - 2)^2 - 4 + 3(y - 2)^2 - 12 + 28 = 0 )Combine constants:-4 -12 +28 = 12So, ( (x - 2)^2 + 3(y - 2)^2 + 12 = 0 )Which simplifies to ( (x - 2)^2 + 3(y - 2)^2 = -12 )Yes, that's correct. Since the sum of squares equals a negative number, there are no real points (x, y) that satisfy this equation. Therefore, the graph is empty.I think that's the conclusion. It's not a parabola, circle, ellipse, hyperbola, point, line, or two lines. It's just empty because there are no real solutions.

💡To solve for ( z ) in the equation ( frac{12}{20} = cbrt{frac{z}{20}} ), follow these steps:1. **Simplify the fraction on the left side:** [ frac{12}{20} = frac{3}{5} = 0.6 ] 2. **Cube both sides of the equation to eliminate the cube root:** [ (0.6)^3 = left(cbrt{frac{z}{20}}right)^3 ] [ 0.216 = frac{z}{20} ] 3. **Solve for ( z ) by multiplying both sides by 20:** [ z = 0.216 times 20 = 4.32 ]

💡Okay, so I have this problem where there's a pot with a diameter of 36 cm. Inside this pot, there are already two jars: one with a radius of 6 cm and another with a radius of 12 cm. I need to figure out the maximum radius of an additional jar that can fit alongside these two. Hmm, let me try to visualize this.First, the pot has a diameter of 36 cm, so its radius is half of that, which is 18 cm. That means the pot can hold jars up to a radius of 18 cm, but since there are already two jars inside, the additional jar has to fit around them without overlapping.Let me think about the sizes of the jars. The first jar has a radius of 6 cm, and the second one is bigger, with a radius of 12 cm. So, the total space taken by these two jars isn't just their radii; it's also the space between them. I guess I need to consider how these jars are arranged inside the pot.If I imagine the pot as a circle with a radius of 18 cm, and the two jars as smaller circles inside it, the additional jar will have to fit in the remaining space. I think the key here is to figure out how much space is left between the two existing jars and the edge of the pot.Maybe I can use some geometry here. If I consider the centers of the jars, the distance from the center of the pot to the center of each jar plus the jar's radius should equal the pot's radius. So, for the first jar with radius 6 cm, the distance from the pot's center to its center would be 18 cm minus 6 cm, which is 12 cm. Similarly, for the second jar with radius 12 cm, the distance from the pot's center to its center would be 18 cm minus 12 cm, which is 6 cm.Wait, that doesn't seem right. If the second jar has a radius of 12 cm, and the pot has a radius of 18 cm, the center of the second jar would actually be 18 cm minus 12 cm, which is 6 cm away from the pot's center. That makes sense because the jar is quite large.Now, the first jar is smaller, with a radius of 6 cm, so its center is 18 cm minus 6 cm, which is 12 cm away from the pot's center. So, the two jars are 12 cm and 6 cm away from the center of the pot, respectively.I think I need to find the distance between the centers of the two existing jars. If I can find that, I can figure out how much space is left for the additional jar. The distance between the centers of the two jars would be the distance between the two points that are 12 cm and 6 cm away from the center of the pot, respectively.Let me represent this with coordinates. Let's place the center of the pot at the origin (0,0). Let's assume the second jar (radius 12 cm) is at (6,0), since it's 6 cm away from the center. The first jar (radius 6 cm) is 12 cm away from the center. To simplify, let's place it at (12,0). Wait, but if both jars are on the same line, their centers are 6 cm apart. But their radii are 12 cm and 6 cm, so the distance between their edges would be 6 cm minus (12 cm + 6 cm). Wait, that doesn't make sense.Hold on, maybe I should use the Pythagorean theorem here. If the two jars are placed such that their centers are at (6,0) and (12,0), the distance between their centers is 6 cm. The sum of their radii is 12 cm + 6 cm = 18 cm. But the distance between their centers is only 6 cm, which is less than the sum of their radii. That means they would overlap. But in reality, they can't overlap because they are separate jars inside the pot.Hmm, maybe my initial placement is incorrect. Perhaps the jars are not aligned along the same diameter. Maybe they are placed at some angle relative to each other. Let me think about that.If the two jars are placed symmetrically around the center, their centers would form an isosceles triangle with the center of the pot. The distance from the center of the pot to each jar's center is 6 cm and 12 cm, respectively. Wait, no, actually, the distance from the center of the pot to the center of the second jar is 6 cm, and to the first jar is 12 cm. So, the two centers are 6 cm and 12 cm away from the origin.To find the distance between the two centers, I can use the law of cosines if I know the angle between them. But I don't know the angle. Maybe I can assume they are placed such that the additional jar can fit in the remaining space. Alternatively, perhaps I can model this as a circle packing problem.In circle packing, we try to fit multiple circles inside a larger circle without overlapping. In this case, we have three circles inside a larger circle. The radii of the three smaller circles are 6 cm, 12 cm, and the unknown radius r. The radius of the larger circle is 18 cm.There's a formula or method to find the maximum radius of the third circle that can fit alongside the two existing ones. I think it involves solving a system of equations based on the distances between the centers.Let me denote the centers of the jars as follows:- Center of the pot: O- Center of the jar with radius 12 cm: O1- Center of the jar with radius 6 cm: O2- Center of the unknown jar: O3The distances from the center of the pot to each jar's center are:- OO1 = 18 - 12 = 6 cm- OO2 = 18 - 6 = 12 cm- OO3 = 18 - rThe distances between the centers of the jars should be equal to the sum of their radii if they are just touching each other:- O1O2 = 12 + 6 = 18 cm- O1O3 = 12 + r- O2O3 = 6 + rNow, we have a triangle formed by O1, O2, and O3. The sides of this triangle are 18 cm, (12 + r) cm, and (6 + r) cm. Also, the distances from the center of the pot to each jar's center are 6 cm, 12 cm, and (18 - r) cm.This seems complicated, but maybe I can set up some equations using the distances. Let me consider the coordinates again. Let's place O1 at (6,0) and O2 somewhere in the plane. The distance between O1 and O2 should be 18 cm. Wait, but O1 is at (6,0), and O2 is 12 cm away from the origin. So, the distance between O1 and O2 can be found using the distance formula.Let me denote O2 as (x,y). Then, the distance from O to O2 is sqrt(x^2 + y^2) = 12 cm. The distance from O1 to O2 is sqrt((x - 6)^2 + y^2) = 18 cm.So, we have two equations:1. x^2 + y^2 = 1442. (x - 6)^2 + y^2 = 324Subtracting equation 1 from equation 2:(x - 6)^2 + y^2 - x^2 - y^2 = 324 - 144Expanding (x - 6)^2: x^2 - 12x + 36 + y^2 - x^2 - y^2 = 180Simplify: -12x + 36 = 180-12x = 144x = -12So, x is -12. Plugging back into equation 1:(-12)^2 + y^2 = 144144 + y^2 = 144y^2 = 0y = 0So, O2 is at (-12, 0). That makes sense because it's 12 cm away from the origin in the opposite direction of O1, which is at (6,0). So, the two jars are placed diametrically opposite each other along the x-axis.Now, where can we place the third jar, O3, such that it touches both O1 and O2 and also fits within the pot? Let's denote O3 as (a,b). The distance from O to O3 is 18 - r, so:sqrt(a^2 + b^2) = 18 - rThe distance from O1 to O3 is 12 + r:sqrt((a - 6)^2 + b^2) = 12 + rThe distance from O2 to O3 is 6 + r:sqrt((a + 12)^2 + b^2) = 6 + rSo, we have three equations:1. a^2 + b^2 = (18 - r)^22. (a - 6)^2 + b^2 = (12 + r)^23. (a + 12)^2 + b^2 = (6 + r)^2Let me expand equations 2 and 3.Expanding equation 2:(a - 6)^2 + b^2 = a^2 - 12a + 36 + b^2 = (12 + r)^2 = 144 + 24r + r^2But from equation 1, a^2 + b^2 = (18 - r)^2 = 324 - 36r + r^2So, substituting a^2 + b^2 from equation 1 into equation 2:324 - 36r + r^2 - 12a + 36 = 144 + 24r + r^2Simplify:324 - 36r + r^2 - 12a + 36 - 144 - 24r - r^2 = 0(324 + 36 - 144) + (-36r - 24r) + (-12a) + (r^2 - r^2) = 0216 - 60r - 12a = 0Divide by 12:18 - 5r - a = 0So, a = 18 - 5rSimilarly, let's expand equation 3:(a + 12)^2 + b^2 = a^2 + 24a + 144 + b^2 = (6 + r)^2 = 36 + 12r + r^2Again, using equation 1, a^2 + b^2 = 324 - 36r + r^2Substitute into equation 3:324 - 36r + r^2 + 24a + 144 = 36 + 12r + r^2Simplify:324 - 36r + r^2 + 24a + 144 - 36 - 12r - r^2 = 0(324 + 144 - 36) + (-36r - 12r) + 24a + (r^2 - r^2) = 0432 - 48r + 24a = 0Divide by 24:18 - 2r + a = 0So, a = 2r - 18Now, from equation 2, we have a = 18 - 5rFrom equation 3, we have a = 2r - 18Set them equal:18 - 5r = 2r - 1818 + 18 = 2r + 5r36 = 7rr = 36/7 ≈ 5.14 cmSo, the maximum radius of the additional jar is 36/7 cm, which is approximately 5.14 cm.Let me just check if this makes sense. If the additional jar has a radius of about 5.14 cm, its center would be 18 - 5.14 ≈ 12.86 cm away from the center of the pot. The distance from O1 (6 cm from center) to O3 should be 12 + 5.14 ≈ 17.14 cm. Let's see:Using a = 18 - 5r = 18 - 5*(36/7) = 18 - 180/7 = (126 - 180)/7 = -54/7 ≈ -7.71 cmSo, O3 is at (-54/7, b). Let's find b using equation 1:a^2 + b^2 = (18 - r)^2(-54/7)^2 + b^2 = (18 - 36/7)^2(2916/49) + b^2 = (126/7 - 36/7)^2(2916/49) + b^2 = (90/7)^2(2916/49) + b^2 = 8100/49b^2 = 8100/49 - 2916/49 = (8100 - 2916)/49 = 5184/49b = sqrt(5184/49) = 72/7 ≈ 10.29 cmSo, O3 is at (-54/7, 72/7). Let's check the distance from O1 to O3:O1 is at (6,0), O3 is at (-54/7, 72/7)Distance = sqrt[(6 + 54/7)^2 + (0 - 72/7)^2]Convert 6 to 42/7:sqrt[(42/7 + 54/7)^2 + (-72/7)^2] = sqrt[(96/7)^2 + (72/7)^2]= sqrt[(9216/49) + (5184/49)] = sqrt[14400/49] = 120/7 ≈ 17.14 cmWhich is equal to 12 + 5.14 ≈ 17.14 cm, so that checks out.Similarly, distance from O2 to O3:O2 is at (-12,0), O3 is at (-54/7, 72/7)Distance = sqrt[(-12 + 54/7)^2 + (0 - 72/7)^2]Convert -12 to -84/7:sqrt[(-84/7 + 54/7)^2 + (-72/7)^2] = sqrt[(-30/7)^2 + (72/7)^2]= sqrt[(900/49) + (5184/49)] = sqrt[6084/49] = 78/7 ≈ 11.14 cmWhich should be equal to 6 + 5.14 ≈ 11.14 cm, so that also checks out.Therefore, the calculations seem consistent. The maximum radius of the additional jar is 36/7 cm, which is approximately 5.14 cm.

💡Okay, so I have this problem here: Ashley secured 332 marks in an exam, and that's 83% of the total marks. I need to find out what the maximum marks were. Hmm, percentages can be tricky sometimes, but I think I can figure this out.First, I remember that percentages are parts per hundred. So, 83% means 83 per 100, or 0.83 in decimal form. That might be useful later. Now, Ashley scored 332 marks, and that's 83% of the total marks. So, I guess I can set up an equation where 332 is equal to 83% of the maximum marks.Let me write that down: 332 = 83% of Maximum Marks. To make this easier, I can convert the percentage to a decimal. So, 83% becomes 0.83. Now, the equation looks like 332 = 0.83 * Maximum Marks.I need to find the Maximum Marks, so I should solve for that. To isolate Maximum Marks, I can divide both sides of the equation by 0.83. That should give me Maximum Marks = 332 / 0.83.Let me do that division. 332 divided by 0.83. Hmm, dividing by a decimal can be a bit confusing. Maybe I can make it easier by multiplying both the numerator and the denominator by 100 to eliminate the decimal. So, that would make it 33200 divided by 83.Now, I need to calculate 33200 divided by 83. I think I can do this by long division. Let's see, how many times does 83 go into 332? Well, 83 times 4 is 332, so that's exactly 4 times. So, 83 goes into 332 four times, and then I have no remainder.Wait, but I'm dividing 33200 by 83. So, after dividing 332 by 83 to get 4, I still have two more zeros to consider. So, it's like 4 times 100, which is 400. So, 33200 divided by 83 is 400.Therefore, the maximum marks must be 400. Let me double-check that to make sure I didn't make a mistake. If 83% of 400 is 332, then 0.83 * 400 should equal 332. Calculating that, 0.83 * 400 is indeed 332. So, that checks out.I think I've got it right. The maximum marks in the exam were 400.

💡Alright, so I have this problem here: "A triangle with integral sides is isosceles and has a perimeter of 12. What is the area of the triangle?" The options are given as (A) 3√3, (B) 4√3, (C) 6, (D) 9√3, and (E) 8. Hmm, okay, let's try to figure this out step by step.First, let's recall what an isosceles triangle is. It's a triangle with at least two sides equal in length. Since the triangle has integral sides, all three sides must be integers. The perimeter is 12, which means the sum of all three sides is 12.Let me denote the lengths of the sides. Since it's isosceles, let's say two sides are equal, so let's call them both 'a' and the third side 'b'. So, the sides are a, a, and b. Therefore, the perimeter is 2a + b = 12.Now, I need to find integer values of a and b such that 2a + b = 12. Also, since it's a triangle, it has to satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the remaining side.So, let's list possible integer values for a and b.Starting with a = 1: Then b = 12 - 2(1) = 10. But then, checking the triangle inequality: 1 + 1 > 10? 2 > 10? No, that's not true. So, a = 1 is invalid.a = 2: Then b = 12 - 4 = 8. Checking triangle inequality: 2 + 2 > 8? 4 > 8? No, still not valid.a = 3: Then b = 12 - 6 = 6. Checking triangle inequality: 3 + 3 > 6? 6 > 6? No, that's not true either because it's not greater, just equal. So, a = 3 is invalid.a = 4: Then b = 12 - 8 = 4. Checking triangle inequality: 4 + 4 > 4? 8 > 4? Yes, that's true. Also, 4 + 4 > 4 is true, and 4 + 4 > 4 is true. So, a = 4 and b = 4 is valid. Wait, but if a = 4 and b = 4, then all sides are equal, making it an equilateral triangle. Hmm, but the problem says it's isosceles, which technically includes equilateral triangles since they have at least two sides equal. So, that's acceptable.a = 5: Then b = 12 - 10 = 2. Checking triangle inequality: 5 + 5 > 2? 10 > 2? Yes. 5 + 2 > 5? 7 > 5? Yes. 5 + 2 > 5? 7 > 5? Yes. So, a = 5 and b = 2 is also valid.a = 6: Then b = 12 - 12 = 0. But a side length can't be zero, so that's invalid.So, the possible integer solutions are a = 4, b = 4 and a = 5, b = 2.Now, let's consider these two cases.First case: a = 4, b = 4. This is an equilateral triangle with all sides equal to 4. The area of an equilateral triangle can be calculated using the formula:Area = (√3 / 4) * side²So, plugging in side = 4:Area = (√3 / 4) * 16 = 4√3Second case: a = 5, b = 2. This is an isosceles triangle with two sides of length 5 and one side of length 2. To find the area, we can use Heron's formula, which is:Area = √[s(s - a)(s - b)(s - c)]where s is the semi-perimeter, which is 12 / 2 = 6.So, plugging in the values:Area = √[6(6 - 5)(6 - 5)(6 - 2)] = √[6 * 1 * 1 * 4] = √[24] = 2√6Hmm, 2√6 is approximately 4.899, which isn't one of the answer choices. Wait, the answer choices are 3√3 (~5.196), 4√3 (~6.928), 6, 9√3 (~15.588), and 8. So, 2√6 isn't among them. That suggests that maybe this case isn't valid or perhaps I made a mistake.Wait, let's double-check the triangle inequality for a = 5 and b = 2. 5 + 5 > 2? 10 > 2? Yes. 5 + 2 > 5? 7 > 5? Yes. 5 + 2 > 5? 7 > 5? Yes. So, it is a valid triangle. But the area isn't matching any of the options. Hmm.Alternatively, maybe I should calculate the area differently. For an isosceles triangle, another way to find the area is to drop a height from the apex to the base, splitting the base into two equal parts. Then, the area is (base * height) / 2.In the case of a = 5, b = 2, the base is 2, and the equal sides are 5 each. So, if we drop a height h from the apex to the base, it will split the base into two segments of length 1 each. Then, using the Pythagorean theorem:h² + 1² = 5²h² + 1 = 25h² = 24h = √24 = 2√6So, the area is (base * height) / 2 = (2 * 2√6) / 2 = 2√6, which is the same as before. So, that's correct.But since 2√6 isn't one of the answer choices, maybe this case isn't considered? Or perhaps the problem expects only non-equilateral isosceles triangles? Wait, the problem says "isosceles," which includes equilateral, but maybe in the context of the problem, they consider only triangles with exactly two sides equal. That might be a possibility.If that's the case, then the only valid case is a = 5, b = 2, but as we saw, the area is 2√6, which isn't an option. Hmm, confusing.Wait, let's check the first case again. If a = 4, b = 4, it's an equilateral triangle with area 4√3, which is option (B). That is one of the answer choices. So, perhaps the problem expects us to consider equilateral triangles as isosceles, and thus the answer is 4√3.But just to be thorough, let's see if there are any other possible integer solutions. We had a = 4 and a = 5 as the only valid ones. a = 3 gives b = 6, but that doesn't satisfy the triangle inequality because 3 + 3 is not greater than 6. Similarly, a = 2 and a = 1 don't work.So, the only valid cases are a = 4, b = 4 and a = 5, b = 2. Since the area for a = 4, b = 4 is 4√3, which is an option, and the other case doesn't give an option, perhaps the answer is 4√3.Alternatively, maybe I made a mistake in considering the triangle with sides 5,5,2. Let me double-check the area calculation.Using Heron's formula:s = (5 + 5 + 2)/2 = 6Area = √[6(6 - 5)(6 - 5)(6 - 2)] = √[6 * 1 * 1 * 4] = √24 = 2√6Yes, that's correct. So, 2√6 is approximately 4.899, which isn't one of the options. Therefore, perhaps the problem expects us to consider only the equilateral triangle case, which gives 4√3, an option.Alternatively, maybe I missed another possible integer solution. Let's see:We had a = 4, b = 4 and a = 5, b = 2. Are there any other integer values for a and b?Wait, if a = 3, b = 6, but as we saw, 3 + 3 is not greater than 6, so that's invalid.a = 6, b = 0, which is invalid.a = 7, b would be negative, which is invalid.So, no, only a = 4 and a = 5 are valid.Therefore, the only possible area that matches the answer choices is 4√3, which is option (B).But just to make sure, let's think again: the problem says "a triangle with integral sides is isosceles and has a perimeter of 12." So, it's possible that the triangle is equilateral, which is a special case of isosceles. Therefore, the area is 4√3.Alternatively, if the problem expects only non-equilateral isosceles triangles, then perhaps there's a mistake in the problem or the answer choices. But since 4√3 is an option and it's a valid area for an equilateral triangle, which is isosceles, I think that's the intended answer.So, I think the answer is (B) 4√3.

💡First, calculate the commission on the first 500 of the sale, which is 20% of 500.Commission on first 500 = 0.20 * 500 = 100Next, determine the amount in excess of 500 from the total sale of 800.Amount in excess of 500 = 800 - 500 = 300Calculate the commission on the excess amount, which is 25% of 300.Commission on excess amount = 0.25 * 300 = 75Add the two commissions to find the total commission.Total commission = 100 + 75 = 175Finally, calculate the commission as a percentage of the total sale.Commission percentage = (175 / 800) * 100 = 21.875%