💡Alright, so I have this problem about a rhombus, and I need to find the lengths of its diagonals. Let me try to visualize the problem first. A rhombus is a four-sided shape where all sides are equal in length, and opposite sides are parallel. The angles can be acute or obtuse, but in this case, we're dealing with an acute angle.The problem says that from the vertex of an acute angle, perpendiculars are drawn to the lines containing the sides of the rhombus, which do not include this vertex. Each of these perpendiculars is 3 cm long, and the distance between their bases is (3sqrt{3}) cm. I need to find the lengths of the diagonals.Okay, let's break this down. First, let's imagine the rhombus. Let's label the rhombus as ABCD, with vertex A being the acute angle from which the perpendiculars are drawn. So, from point A, we draw perpendiculars to the sides that do not include A. Since it's a rhombus, sides AB and AD include vertex A, so the perpendiculars must be drawn to sides BC and CD.Wait, actually, the problem says "the lines containing the sides of the rhombus, which do not include this vertex." So, if we're at vertex A, the sides that include A are AB and AD, so the sides that do not include A are BC and CD. Therefore, the perpendiculars are drawn from A to lines BC and CD.But in a rhombus, sides BC and CD are adjacent to vertex C, not A. So, drawing a perpendicular from A to BC and another from A to CD. These perpendiculars will meet BC and CD at some points, say E and F respectively. The lengths of these perpendiculars are both 3 cm, and the distance between points E and F is (3sqrt{3}) cm.Hmm, okay. So, we have two perpendiculars from A to BC and CD, each 3 cm long, and the distance between their bases (E and F) is (3sqrt{3}) cm. I need to relate this information to find the diagonals of the rhombus.Let me recall that in a rhombus, the diagonals bisect each other at right angles. Also, the diagonals bisect the angles of the rhombus. So, if I can find the lengths of the diagonals, I can use properties related to the angles and sides.Since the perpendiculars from A to BC and CD are both 3 cm, maybe I can use some trigonometric relationships here. Let me consider the angles at vertex A. Let’s denote the acute angle at A as θ. Then, the obtuse angle at C would be 180° - θ.In a rhombus, the sides are equal, so all sides are of length, let's say, 'a'. The diagonals can be expressed in terms of the sides and the angles. Specifically, the lengths of the diagonals are (2asin(theta/2)) and (2acos(theta/2)). Wait, is that correct?Actually, the diagonals can be found using the formulas:- The length of the shorter diagonal (which is opposite the acute angle) is (2asin(theta/2)).- The length of the longer diagonal (which is opposite the obtuse angle) is (2acos(theta/2)).But I'm not sure if that's the exact formula. Maybe I should derive it.In a rhombus, the diagonals split the angles into two equal parts. So, if the acute angle at A is θ, then each half-angle is θ/2. The diagonals intersect at right angles, so we can consider the triangles formed by the diagonals.Let me denote the diagonals as AC and BD, intersecting at point O, which is the midpoint of both diagonals. Then, triangles AOB, BOC, COD, and DOA are all right-angled triangles.In triangle AOB, angle OAB is θ/2, and sides OA and OB are half the lengths of the diagonals AC and BD, respectively. So, if I denote OA = (AC)/2 and OB = (BD)/2, then in triangle AOB, we have:[sin(theta/2) = frac{OB}{AB} = frac{(BD)/2}{a}][cos(theta/2) = frac{OA}{AB} = frac{(AC)/2}{a}]So, from these, we can express the diagonals in terms of 'a' and θ:[BD = 2asin(theta/2)][AC = 2acos(theta/2)]Okay, so that seems correct. Now, going back to the problem. We have perpendiculars from A to BC and CD, each of length 3 cm. Let me think about how to relate this to the angles and sides.Since we're dealing with perpendiculars from A to BC and CD, these are essentially the heights from A to those sides. In a rhombus, the height can be expressed as (asin(theta)), where θ is the acute angle.Wait, is that right? Let me recall that the area of a rhombus can be expressed as (a^2sin(theta)), and it can also be expressed as base times height. So, if we take one side as the base, the height would be (asin(theta)). So, yes, the height from A to BC would be (asin(theta)), and similarly, the height from A to CD would also be (asin(theta)).But in this problem, both heights are given as 3 cm. So, that suggests that (asin(theta) = 3). So, we have one equation: (asin(theta) = 3).Now, the distance between the bases of these perpendiculars is (3sqrt{3}) cm. So, the distance between points E and F is (3sqrt{3}) cm.I need to relate this distance to the geometry of the rhombus. Let me try to visualize or sketch the rhombus with the perpendiculars drawn.So, from point A, we draw two perpendiculars: one to BC, meeting at E, and another to CD, meeting at F. The distance between E and F is (3sqrt{3}) cm.Since BC and CD are sides of the rhombus, and E and F are points on BC and CD respectively, the line segment EF is inside the rhombus, connecting these two points.I need to find the length of EF, which is (3sqrt{3}) cm, in terms of the sides and angles of the rhombus.Perhaps I can consider triangle AEF. Since AE and AF are both 3 cm, and EF is (3sqrt{3}) cm, triangle AEF is an isosceles triangle with sides AE = AF = 3 cm and base EF = (3sqrt{3}) cm.Wait, is triangle AEF a triangle within the rhombus? Let me think. Point E is on BC, and point F is on CD. So, connecting E and F would create a line segment inside the rhombus, but is triangle AEF a valid triangle?Actually, point A is connected to E and F, so triangle AEF is formed by points A, E, and F. Since AE and AF are both perpendiculars from A, and E and F are on BC and CD respectively, triangle AEF is indeed a triangle inside the rhombus.Given that, triangle AEF has sides AE = AF = 3 cm and EF = (3sqrt{3}) cm. So, it's an isosceles triangle with two sides equal to 3 cm and the base equal to (3sqrt{3}) cm.I can use the properties of this triangle to find some angles or other lengths. Maybe I can find the angle at A in triangle AEF, which might relate to the angle θ of the rhombus.Let me calculate the angles of triangle AEF. Since it's an isosceles triangle with sides AE = AF = 3 cm and base EF = (3sqrt{3}) cm, I can use the Law of Cosines to find the angle at A.The Law of Cosines states that for any triangle with sides a, b, and c, opposite angles A, B, and C respectively:[c^2 = a^2 + b^2 - 2abcos(C)]In triangle AEF, let's denote angle at A as α. Then, sides AE = AF = 3 cm, and EF = (3sqrt{3}) cm. Applying the Law of Cosines:[(3sqrt{3})^2 = 3^2 + 3^2 - 2 times 3 times 3 times cos(alpha)][27 = 9 + 9 - 18cos(alpha)][27 = 18 - 18cos(alpha)][27 - 18 = -18cos(alpha)][9 = -18cos(alpha)][cos(alpha) = -frac{9}{18} = -frac{1}{2}]So, angle α is arccos(-1/2), which is 120°. Therefore, the angle at A in triangle AEF is 120°.But in the rhombus, the angle at A is θ, which is acute. So, how does this 120° angle relate to θ?Wait, maybe angle α is related to θ. Since triangle AEF is inside the rhombus, and angle at A in the rhombus is θ, perhaps angle α is supplementary to θ? Because if I consider the direction of the perpendiculars, the angle between AE and AF might be related to the angle of the rhombus.Let me think. In the rhombus, sides AB and AD meet at angle θ. The perpendiculars from A to BC and CD are AE and AF, respectively. So, the angle between AE and AF is actually the angle between two lines that are each perpendicular to sides BC and CD.Since BC and CD are adjacent sides of the rhombus meeting at C, the angle between BC and CD is the obtuse angle, which is 180° - θ. Therefore, the angle between the perpendiculars AE and AF would be equal to the angle between BC and CD, which is 180° - θ.But in triangle AEF, we found that angle α is 120°, which is the angle between AE and AF. Therefore, 180° - θ = 120°, so θ = 60°.Ah, so the acute angle θ at vertex A is 60°, and the obtuse angle at vertex C is 120°.Now that we know θ = 60°, we can relate this back to the sides and diagonals of the rhombus.Earlier, we had the equation (asin(theta) = 3). Since θ = 60°, we can substitute:[asin(60°) = 3][a times frac{sqrt{3}}{2} = 3][a = 3 times frac{2}{sqrt{3}} = frac{6}{sqrt{3}} = 2sqrt{3} text{ cm}]So, the side length 'a' of the rhombus is (2sqrt{3}) cm.Now, we can find the lengths of the diagonals using the formulas I derived earlier:[BD = 2asin(theta/2)][AC = 2acos(theta/2)]Given that θ = 60°, let's compute these.First, compute BD:[BD = 2 times 2sqrt{3} times sin(30°)][BD = 4sqrt{3} times frac{1}{2}][BD = 2sqrt{3} text{ cm}]Next, compute AC:[AC = 2 times 2sqrt{3} times cos(30°)][AC = 4sqrt{3} times frac{sqrt{3}}{2}][AC = 4sqrt{3} times frac{sqrt{3}}{2} = 4 times frac{3}{2} = 6 text{ cm}]So, the lengths of the diagonals BD and AC are (2sqrt{3}) cm and 6 cm, respectively.Let me just verify if these results make sense. The side length is (2sqrt{3}) cm, and the diagonals are (2sqrt{3}) cm and 6 cm. In a rhombus, the diagonals should satisfy the relationship:[text{Area} = frac{d_1 times d_2}{2}][text{Area} = a^2 sin(theta)]Let's compute both:First, using the diagonals:[text{Area} = frac{2sqrt{3} times 6}{2} = frac{12sqrt{3}}{2} = 6sqrt{3} text{ cm}^2]Using the side length and angle:[text{Area} = (2sqrt{3})^2 times sin(60°) = 12 times frac{sqrt{3}}{2} = 6sqrt{3} text{ cm}^2]Both methods give the same area, so that checks out. Also, since the diagonals are perpendicular bisectors, they form right triangles with the sides. Let's check if the Pythagorean theorem holds for the triangles formed by the diagonals.Each half of diagonal BD is (sqrt{3}) cm, and each half of diagonal AC is 3 cm. So, in triangle AOB, sides OA = 3 cm, OB = (sqrt{3}) cm, and hypotenuse AB = (2sqrt{3}) cm.Applying Pythagoras:[OA^2 + OB^2 = AB^2][3^2 + (sqrt{3})^2 = (2sqrt{3})^2][9 + 3 = 12][12 = 12]That works out. So, all the relationships are consistent.Therefore, the lengths of the diagonals are (2sqrt{3}) cm and 6 cm.

💡First, I need to determine the number of supervisors and laborers in the workshop. I'll start by calculating the total salary of the technicians, which is 6 multiplied by Rs. 12,000, giving Rs. 72,000.Next, I'll set up an equation for the total salary of the supervisors and laborers combined, which is given as Rs. 270,000. This can be expressed as 15,000 times the number of supervisors plus 6,000 times the number of laborers.Then, I'll calculate the total salary of all workers by adding the salary of the technicians to the combined salary of the supervisors and laborers, resulting in Rs. 342,000.Since the average salary of all workers is Rs. 9,000, I'll divide the total salary by this average to find the total number of workers. Dividing Rs. 342,000 by Rs. 9,000 gives me 38.Therefore, the total number of workers in the workshop is 38.

💡Okay, so I have this problem where several straight lines are drawn on a plane, and each line intersects exactly 15 other lines. I need to figure out how many lines are drawn in total. Hmm, let me think about how to approach this.First, I remember that when dealing with lines on a plane, two lines can either intersect or be parallel. If they're parallel, they don't intersect, right? So, if each line intersects exactly 15 others, that means each line is not parallel to 15 other lines. But how does that help me find the total number of lines?Maybe I should consider the concept of equivalence classes. If I define an equivalence relation where two lines are equivalent if they are parallel, then the set of all lines is partitioned into classes where each class consists of lines that are all parallel to each other. Let's say there are ( n ) such classes, and each class has ( a_1, a_2, ldots, a_n ) lines respectively. So, the total number of lines ( S ) would be the sum of all these ( a_i )s: ( S = a_1 + a_2 + ldots + a_n ).Now, each line intersects exactly 15 other lines. Since lines within the same class are parallel, they don't intersect each other. Therefore, each line must intersect all the lines in the other classes. So, for a line in class ( a_i ), the number of lines it intersects is ( S - a_i ). According to the problem, this number is 15. So, for each class ( a_i ), we have:[S - a_i = 15]Which implies:[a_i = S - 15]Wait, that's interesting. So, every class has the same number of lines, ( S - 15 ). But how many classes are there? Let me denote the number of classes as ( n ). Since each class has ( a_i = S - 15 ) lines, the total number of lines ( S ) can be expressed as:[S = n times (S - 15)]Let me write that equation out:[S = n(S - 15)]Simplifying this, I get:[S = nS - 15n][S - nS = -15n][S(1 - n) = -15n][S = frac{-15n}{1 - n}][S = frac{15n}{n - 1}]Hmm, so ( S ) must be equal to ( frac{15n}{n - 1} ). Since ( S ) is the total number of lines, it must be a positive integer. Therefore, ( frac{15n}{n - 1} ) must be an integer. Let me denote ( k = n - 1 ), so ( n = k + 1 ). Substituting back, we get:[S = frac{15(k + 1)}{k}][S = 15 + frac{15}{k}]So, ( frac{15}{k} ) must be an integer, which means ( k ) must be a positive divisor of 15. The positive divisors of 15 are 1, 3, 5, and 15. Therefore, ( k ) can be 1, 3, 5, or 15, which means ( n = k + 1 ) can be 2, 4, 6, or 16.Now, let's compute ( S ) for each possible ( k ):1. If ( k = 1 ), then ( n = 2 ) and ( S = 15 + frac{15}{1} = 30 ).2. If ( k = 3 ), then ( n = 4 ) and ( S = 15 + frac{15}{3} = 20 ).3. If ( k = 5 ), then ( n = 6 ) and ( S = 15 + frac{15}{5} = 18 ).4. If ( k = 15 ), then ( n = 16 ) and ( S = 15 + frac{15}{15} = 16 ).So, the possible values for ( S ) are 16, 18, 20, and 30. Let me verify if these make sense.- For ( S = 16 ): There are 16 classes, each with ( 16 - 15 = 1 ) line. So, 16 lines, each intersecting the other 15. That works because each line is in its own class, so they all intersect each other. - For ( S = 18 ): There are 6 classes, each with ( 18 - 15 = 3 ) lines. So, 6 sets of 3 parallel lines. Each line intersects the 15 lines from the other 5 classes. That makes sense.- For ( S = 20 ): There are 4 classes, each with ( 20 - 15 = 5 ) lines. Each line intersects the 15 lines from the other 3 classes. That also works.- For ( S = 30 ): There are 2 classes, each with ( 30 - 15 = 15 ) lines. Each line intersects the 15 lines from the other class. Perfect.So, all these possibilities satisfy the condition that each line intersects exactly 15 others. Therefore, the total number of lines can be 16, 18, 20, or 30.I think that covers all the cases because we considered all the divisors of 15 for ( k ), which gave us all possible values for ( n ) and consequently for ( S ). I don't see any other possibilities here, so I'm confident these are the only solutions.**Final Answer**The number of lines drawn on the plane can be boxed{16}, boxed{18}, boxed{20}, or boxed{30}.

💡Alright, so I've got this problem about a function y = f(x) and two points A and B on its graph. The problem defines something called "curvature" between these two points, which is denoted by φ(A, B). It's calculated as the absolute difference of the slopes of the tangents at A and B divided by the distance between A and B. That is, φ(A, B) = |k_A - k_B| / |AB|, where k_A and k_B are the slopes of the tangents at A and B, and |AB| is the length of the segment connecting A and B.There are four statements given, and I need to determine which ones are correct. Let me go through each one step by step.**Statement 1:** For the function y = x³, if the x-coordinates of points A and B are 1 and -1 respectively, then φ(A, B) = 0.Okay, so let's break this down. First, I need to find the slopes of the tangents at points A and B. The function is y = x³, so its derivative is y' = 3x². At point A, where x = 1, the slope k_A = 3*(1)² = 3.At point B, where x = -1, the slope k_B = 3*(-1)² = 3.So, both slopes are equal: k_A = k_B = 3. Therefore, |k_A - k_B| = |3 - 3| = 0.Now, the distance |AB| is the distance between points A and B. Let's find the coordinates of A and B.For A: x = 1, y = (1)³ = 1, so A is (1, 1).For B: x = -1, y = (-1)³ = -1, so B is (-1, -1).The distance between A and B is sqrt[(1 - (-1))² + (1 - (-1))²] = sqrt[(2)² + (2)²] = sqrt[4 + 4] = sqrt[8] = 2*sqrt(2).But since the numerator is 0, φ(A, B) = 0 / (2*sqrt(2)) = 0. So, φ(A, B) is indeed 0. Therefore, statement 1 is correct.**Statement 2:** There exists such a function whose graph has a constant "curvature" between any two points.Hmm, okay. So, we need to find a function where φ(A, B) is constant for any two points A and B on its graph.Let me think about what φ(A, B) represents. It's the absolute difference of the slopes divided by the distance between the points. So, if φ(A, B) is constant, that means |k_A - k_B| / |AB| = constant.Let me denote this constant as c. So, |k_A - k_B| = c * |AB|.I need to find a function f(x) such that for any two points A(x1, f(x1)) and B(x2, f(x2)), the above equation holds.Let me consider linear functions first. For a linear function, the slope is constant everywhere, so k_A = k_B for any A and B. Therefore, |k_A - k_B| = 0, which would make φ(A, B) = 0. So, a linear function would have φ(A, B) = 0, which is constant. But is 0 considered a constant? Yes, it is. So, any linear function would satisfy this condition.Wait, but the problem says "a function whose graph has a constant curvature between any two points." So, if we take a linear function, it's true that φ(A, B) is constant (zero) for any two points. Therefore, such a function exists. So, statement 2 is correct.But wait, maybe the problem is expecting a non-zero constant? Let me think. If we take a function where the curvature is non-zero and constant, is that possible?Wait, in the usual sense, curvature of a curve is a measure of how much it deviates from being a straight line. For a circle, the curvature is constant. But in this problem, the definition of curvature is different. It's |k_A - k_B| / |AB|.So, perhaps a function with constant curvature as defined here might not necessarily be a circle. Let me see.Suppose we have a function where the derivative f'(x) is linear, so f''(x) is constant. Then, the slope k_A - k_B would be proportional to (x1 - x2). Let's see.Let me assume f'(x) = m x + b, so f''(x) = m.Then, |k_A - k_B| = |m x1 + b - (m x2 + b)| = |m (x1 - x2)|.Now, |AB| is the distance between A and B, which is sqrt[(x1 - x2)^2 + (f(x1) - f(x2))^2].If f'(x) is linear, then f(x) is a quadratic function, say f(x) = (m/2) x² + b x + c.Then, f(x1) - f(x2) = (m/2)(x1² - x2²) + b(x1 - x2) = (m/2)(x1 - x2)(x1 + x2) + b(x1 - x2) = (x1 - x2)[(m/2)(x1 + x2) + b].So, |AB| = sqrt[(x1 - x2)^2 + (x1 - x2)^2 * [(m/2)(x1 + x2) + b]^2] = |x1 - x2| * sqrt[1 + [(m/2)(x1 + x2) + b]^2].Therefore, φ(A, B) = |m (x1 - x2)| / [|x1 - x2| * sqrt[1 + [(m/2)(x1 + x2) + b]^2]] = |m| / sqrt[1 + [(m/2)(x1 + x2) + b]^2].Now, for φ(A, B) to be constant, the denominator must be proportional to |m|, but it depends on x1 and x2. So, unless [(m/2)(x1 + x2) + b] is constant for all x1 and x2, which is impossible unless m = 0 and b is constant, which brings us back to a linear function where φ(A, B) = 0.Therefore, the only function where φ(A, B) is constant (and non-zero) would require that [(m/2)(x1 + x2) + b] is constant for all x1 and x2, which is only possible if m = 0, leading back to a linear function with φ(A, B) = 0.Therefore, the only functions where φ(A, B) is constant are linear functions with φ(A, B) = 0. So, statement 2 is correct because such functions exist (linear functions).**Statement 3:** Let A, B be two distinct points on the parabola y = x² + 1, then φ(A, B) ≤ 2.Alright, so the function is y = x² + 1. Let's compute φ(A, B).First, find the slopes at A and B. The derivative is y' = 2x.So, k_A = 2x1 and k_B = 2x2.Therefore, |k_A - k_B| = |2x1 - 2x2| = 2|x1 - x2|.Now, compute |AB|, the distance between A and B.Points A and B are (x1, x1² + 1) and (x2, x2² + 1).So, |AB| = sqrt[(x1 - x2)^2 + (x1² + 1 - (x2² + 1))^2] = sqrt[(x1 - x2)^2 + (x1² - x2²)^2].Simplify (x1² - x2²) = (x1 - x2)(x1 + x2).Therefore, |AB| = sqrt[(x1 - x2)^2 + (x1 - x2)^2(x1 + x2)^2] = |x1 - x2| * sqrt[1 + (x1 + x2)^2].Therefore, φ(A, B) = |k_A - k_B| / |AB| = [2|x1 - x2|] / [|x1 - x2| * sqrt[1 + (x1 + x2)^2]] = 2 / sqrt[1 + (x1 + x2)^2].Now, we need to show that φ(A, B) ≤ 2.Since sqrt[1 + (x1 + x2)^2] ≥ 1 for all real numbers x1 and x2, because (x1 + x2)^2 is always non-negative.Therefore, 2 / sqrt[1 + (x1 + x2)^2] ≤ 2 / 1 = 2.Hence, φ(A, B) ≤ 2. So, statement 3 is correct.**Statement 4:** Let A(x1, y1) and B(x2, y2) be two distinct points on the curve y = e^x, then φ(A, B) < 1.Alright, let's analyze this.The function is y = e^x. Its derivative is y' = e^x.So, k_A = e^{x1} and k_B = e^{x2}.Therefore, |k_A - k_B| = |e^{x1} - e^{x2}|.Now, compute |AB|, the distance between A and B.Points A and B are (x1, e^{x1}) and (x2, e^{x2}).So, |AB| = sqrt[(x1 - x2)^2 + (e^{x1} - e^{x2})^2].Therefore, φ(A, B) = |e^{x1} - e^{x2}| / sqrt[(x1 - x2)^2 + (e^{x1} - e^{x2})^2].We need to show that this is less than 1.Let me denote Δx = x1 - x2 and Δy = e^{x1} - e^{x2}.Then, φ(A, B) = |Δy| / sqrt[(Δx)^2 + (Δy)^2].Notice that sqrt[(Δx)^2 + (Δy)^2] ≥ |Δy|, because sqrt(a² + b²) ≥ |b| for any real numbers a and b.Therefore, |Δy| / sqrt[(Δx)^2 + (Δy)^2] ≤ |Δy| / |Δy| = 1.But we need to show that it's strictly less than 1, i.e., φ(A, B) < 1.This would be true if sqrt[(Δx)^2 + (Δy)^2] > |Δy|, which is equivalent to (Δx)^2 + (Δy)^2 > (Δy)^2, which simplifies to (Δx)^2 > 0.Since A and B are distinct points, x1 ≠ x2, so Δx ≠ 0, hence (Δx)^2 > 0.Therefore, sqrt[(Δx)^2 + (Δy)^2] > |Δy|, which implies that φ(A, B) = |Δy| / sqrt[(Δx)^2 + (Δy)^2] < 1.Hence, statement 4 is correct.**Conclusion:**All four statements are correct.

💡First, I need to understand the relationship between dividing the number by 5 and dividing it by 6. The problem states that when the number is divided by 5, the result is 70 more than when it is divided by 6.I'll represent the unknown number as ( x ).According to the problem, dividing ( x ) by 5 gives a result that is 70 more than dividing ( x ) by 6. This can be written as:[frac{x}{5} = 70 + frac{x}{6}]To solve for ( x ), I'll first eliminate the fractions by finding a common denominator. The least common multiple of 5 and 6 is 30. I'll multiply both sides of the equation by 30 to simplify:[30 times frac{x}{5} = 30 times left(70 + frac{x}{6}right)][6x = 2100 + 5x]Next, I'll isolate ( x ) by subtracting ( 5x ) from both sides:[6x - 5x = 2100][x = 2100]Therefore, the number is 2100.

💡First, I need to determine the width of the rectangle. Let's denote the width as ( w ) inches.According to the problem, the length of the rectangle is 4 times the width, so the length is ( 4w ) inches.The area of a rectangle is calculated by multiplying its length by its width. The problem states that the area is 196 square inches. Therefore, I can set up the equation:[ 196 = w times 4w ]Simplifying this equation:[ 196 = 4w^2 ]To solve for ( w^2 ), I divide both sides of the equation by 4:[ frac{196}{4} = w^2 ][ 49 = w^2 ]Taking the square root of both sides gives:[ w = sqrt{49} ][ w = 7 ]So, the width of the rectangle is 7 inches.