Тема: How do I solve this surface area problem?

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Area is the quantity that expresses the extent of a two-dimensional figure or shape , or planar lamina , in the plane. Surface area is its analog on the two-dimensional surface of a three-dimensional object. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. [1] It is the two-dimensional analog of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept).

The area of a shape can be measured by comparing the shape to squares of a fixed size. [2] In the International System of Units (SI), the standard unit of area is the square metre (written as m 2 ), which is the area of a square whose sides are one metre long. [3] A shape with an area of three square metres would have the same area as three such squares. In mathematics , the unit square is defined to have area one, and the area of any other shape or surface is a dimensionless real number.

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Area is the quantity that expresses the extent of a two-dimensional figure or shape , or planar lamina , in the plane. Surface area is its analog on the two-dimensional surface of a three-dimensional object. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. [1] It is the two-dimensional analog of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept).

The area of a shape can be measured by comparing the shape to squares of a fixed size. [2] In the International System of Units (SI), the standard unit of area is the square metre (written as m 2 ), which is the area of a square whose sides are one metre long. [3] A shape with an area of three square metres would have the same area as three such squares. In mathematics , the unit square is defined to have area one, and the area of any other shape or surface is a dimensionless real number.

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From trevor neale of TF Warren Group on May 8, 2017:
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May 1 - May 5, 2017
Is it practical to use black light to check for dirt or other foreign matter during surface preparation inspection?

save after for extra help 4n+6-2n=2(n+3) simplify formerly than you clean up first simplify to 2n+6=2(n+3) on the grounds which you re taking 2n faraway from the common 4n. Then distribute on the different area using certainty there is not any sign between the two and the parentheses you multiply yet for the clarification that there are parentheses you multiply 2 by applying way of n and 2 by applying way of three so now your equation is 2n+6=2n+6 as a effect n=a million which additionally recommend it has limitless recommendations on the grounds that of the identity belongings. wish this helped(:

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All these problems are done the same way. There is some quantity you are trying to maximize/minimize, and some other quantity you are keeping constant. 1. Write down the equation for the quantity you are trying to maximize/minimize in terms of the free parameter(s). 2. Write down the equation for the quantity which is constant. Rearrange it to get one parameter in terms of the other (generally you have two). 3. Substitute that into equation 1. Now you have a quantity in terms of only one variable. 4. Take the derivative and set it equal to 0. Here s how that works here: 1. "open-tank square base". OK, so it is a 5-sided box. No top. The bottom is square, so the length and width are the same. Call them s. The height is the other free variable. Call it h. The area is the sum of the four sides and the bottom A = s*h + s*h + s*h + s*h + s*s = 4sh + s^2 This is the quantity you re trying to minimize. It has two free parameters as I said. 2. The volume is s * s * h and it is a constant. So s^2 * h = 43.5 That means h = 43.5/s^2. You now have an expression for h in terms of s. 3. Substituting, A = 4s *43.5/s^2 + s^2 = 174/s + s^2 4. Take the derivative with respect to s: dA/ds = -174/s^2 + 2s = 0 Solve for s.

The answer is 21.9474003 yards^2. To do this, we ll split the problem into two parts: the surface area of the base, and the surface area of the sides. To find the surface area of the base, observe that the octagonal base can be split up into 8 equal-sized isosceles triangles, each of whose three vertices consist of (1) two adjacent vertices of the octagon and (2) the center of the octagon. To find the surface area of the sides (i.e., the lateral surface area), observe that there are also 8 equal-sized isosceles triangles forming the sides of the pyramid, each of whose three vertices consist of (1) two adjacent vertices of the octagon, and (2) the tip of the pyramid. (THE BASE) Consider any isosceles triangle in the base. The apothem splits that triangle into two right triangles. The height of the right triangle is the length of the apothem (1.5 yards), and the angle corresponding to the vertex at the center of the octagon is 360 degrees / 16 = 22.5 degrees (since there are 8 isosceles triangles in the octagon, and the apothem splits each isosceles triangle into two identical right triangles, so 8x2 = 16). If we let "b" denote the length of the side of the right triangle that is perpendicular to the apothem, then tan (22.5 degrees) = b / 1.5 yards so b = 1.5 yards x tan (22.5 degrees) = 0.621320344 yards. Therefore, each right triangle has an area of bh/2 = 0.621320344 yards x 1.5 yards / 2 = 0.465990258 yards^2. The base contains 16 of these triangles (2 per isosceles triangle in the octagon), so the total area of the base is A_base = 0.465990258 yards^2 x 16 = 7.45584412 yards^2. (THE SIDES) The sides are done in a similar manner to the base. Again, we ll split each of the isosceles triangles on the sides into two equal-sized right triangles. This time, however, the height of the triangle will no longer be the apothem (1.5 yards) but will instead be the line segment running from the tip of the pyramid to the midpoint of one edge of the base. The "trick" is to figure out the length of this line segment. To do this, observe that this line segment is the hypotenuse of a right triangle, whose other two sides are the height of the pyramid (2.5 yards) and the apothem of the base (1.5 yards). Using the Pythagorean theorem, we have L^2 = (1.5 yards)^2 + (2.5 yards)^2 = 8.5 yards^2 L = 2.91547595 yards. So, the surface area of a right triangle covering half of each side of the pyramid is bL/2 = 0.621320344 yards x 2.91547595 yards / 2 = 0.90572226 yards^2 As before, it follows that the total area of the sides is A_sides = 0.90572226 yards^2 x 16 = 14.4915562 yards^2. Therefore, the total surface area is A = A_base + A_sides = 7.45584412 yards^2 + 14.4915562 yards^2 = 21.9474003 yards^2.

Area is the quantity that expresses the extent of a two-dimensional figure or shape , or planar lamina , in the plane. Surface area is its analog on the two-dimensional surface of a three-dimensional object. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. [1] It is the two-dimensional analog of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept).

The area of a shape can be measured by comparing the shape to squares of a fixed size. [2] In the International System of Units (SI), the standard unit of area is the square metre (written as m 2 ), which is the area of a square whose sides are one metre long. [3] A shape with an area of three square metres would have the same area as three such squares. In mathematics , the unit square is defined to have area one, and the area of any other shape or surface is a dimensionless real number.