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Find the measure, in terms of x, of each side of a square if the Area = x^2– 16x +64.

  Find the measure, in terms of x, of each side of a square if the Area = x^2– 16x +64. Factor and solve. x^2-x-72=0 2x^2+9x-5=0 x^2-64=0 4x^2-36x+72  

Research, Knowledge and Information :


a squares perimeter measures 64 in. if a similar square's ...


a squares perimeter measures 64 in. if a similar square's dimensions are 1/2 ... find out the measure of each side. ... 64. 4x=64. x=16. Now that we know that each ...
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Algebra- 4 Flashcards | Quizlet


Start studying Algebra- 4. Learn vocabulary, terms, ... Liliana decides to crop a square photo 2 inches on each side to ... (x + 2)2 = 121, x represents the side ...
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Geometry (2/2) Flashcards | Quizlet


Geometry (2/2) term (chapter) ... 2) Find length of each side ... fraction of area Area of a sector=(arc's measure/360)*(pi)r^2. Segment ...
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Each side of a square measures 5x2 units. Find the area of ...


Each side of a square measures 5x2 units. Find the area of the square in terms of x.? Find answers now! ... Each side of a square measures 5x2 units.
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Finding Area | Wyzant Resources


We know that the area of a square is length times ... When we combine like terms, we would get 4x + 4 = 64. ... we would divide each side by 4 (to get x by itself) ...
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Square (Geometry) - Math Is Fun


Square (Geometry) (Jump to Area of a Square or Perimeter of a Square) ... Each internal angle is 90 ... The Diagonal is the side length times the square root of 2:
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Triangles - Warrick County School Corporation


EXAMPLE1 Finding an Angle Measure in a Triangle Find the value of x in the ... x 71 Subtract 109 from each side. ... Divide the square into 16 smaller squares with side
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Perimeters and Areas - Andrews University


Perimeters and Areas ... The square is the simplest figure to find the area of (A = x 2, ... (16"-10")/2=3" is the side of a triangle whose other side is the height ...
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Suggested Questions And Answer :


Find the measure, in terms of x, of each side of a square if the Area = x^2– 16x +64.

solushun tu equashun is (x-8)*(x-8)=0, so x=8 if area=8, square side =root(8)=2*root(2)
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find the area of the square in terms of the variable x.

Question: the length of a side of a square is (4x+4) km. find the area of the square in terms of the variable x. The side of the square is s = (4x + 4) The area of a square is A = s^2 So, the area of this square is A = (4x + 4)^2 = 4^2(x+1)^1 = 16(x+1)^2 = 16(x^2 + 2x + 1) Area of square is A = 16x^2 + 32x + 16 km^2 The correct answer is option (b)
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A sheep pen

This answer has now been revised. I hope it's correct according to the given figures. Sorry for the error. I must have confused two very similar questions. The perimeter of the pen is 2L+2W where L and W are length and width. (a) The fence has a length 2L+W because the shed's length is 30m and forms a width. Since W=30, 2L+30=200, 2L=170, L=85m. The area is 30*85=2550 sq m. [corrected] (b) The maximum area for the pen is a square, so L=W and the area is L^2. The perimeter is 4L. We know that the shed helps to form a side. So the perimeter is 200+30=230m=4L, so L=57.5m and the area is 3306.25 sq m. [corrected] It's easy to prove that a square is the maximum area for a given perimeter. If L=a-x and W=a+x, the fixed perimeter, P,  is 4a, so a=P/4. The area is a^2-x^2. This quantity has a maximum value of a^2 only when x=0, which means L=W=a, a square. (c) (i) Now, if the length of each pen is L and the total amount of fencing is 200m, then we can write an equation for the total amount of fencing:  3L+3W=200, where W=30, since the shed covers the width of one pen, so L=110/3=36 2/3m. (ii) If the two pens are square for maximum area each, and the side is L, the total enclosed perimeter is 7L. The perimeter is made up of 200m of fence and 30m of shed=230m. L=230/7=32 6/7m=32.857m approx. [corrected] (d) For three pens of maximum area the perimeter is 10L=200+30=230, and L=23m. The area of each pen is 529 sq m. [corrected] Total pen area is 1587 sq m.
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what is the distance of 800 sqft, 1800sqft and 1250 sqft

Distance is measured in feet (or other linear measurement) and area in square feet (or other measure of area. Linear measurements have one dimension and area measurements have two, so you can't compare them or convert one into the other. A line doesn't enclose anything, because it's just a line. An area is an enclosure like a circle, rectangle, triangle, square, and so on. To make an area out of a line you would have to bend it and join the ends together. The best we can do with the figures in the question is to bend a line to make a shape. Let's go for a 45 degree right-angled triangle. The area of this type of triangle is half the area of a square split into half along a diagonal, and the area of a square is the square of the length of the side, so the area of the triangle is half this value. Now look at the figures representing the area of the triangle. To find the area of the square we simply double the values: 1600, 3600 and 2500. The sides of the corresponding squares would be 40, 60 and 50 (40*40=1600, 60*60=3600 and 50*50=2500). So we know that the sides of each triangle forming the right angle must be 40, 60 or 50. The length of the third side of the triangle is the length of the diagonal of the square and would have respective lengths 56.57ft, 84.85ft and 70.71ft. The length of the lines forming these triangles would therefore be 136.57ft, 204.85ft and 170.71ft. These are the perimeter lengths of the triangles. So now I hope you can see the difference between feet and square feet. We could have bent our line into a circle and the line would be the circumference of the circle and the answers would be quite different. Or we could have made another shape, all giving different lengths for the lines making the shape.
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find the length of each boundary

find the length of each boundary Vista county is setting aside a large parcel of land to preserve it as open space. the county has hired Jane's surveying firm to survey the parcel, which is in the shape of a right triangle. the longer leg of the triangle measures 5 miles less than the square of the shorter leg, and the hypotenuse of the triangle measures 13 miles less than twice the square of the shorter leg. the length of each boundary is a whole number. find the length of each boundary. Make a the long side, b the short side and c the hypotenuse. a = b^2 - 5 b = b, obviously c = 2b^2 - 13 From the Pythagorean theorem, we have a^2 + b^2 = c^2 (b^2 - 5)^2 + b^2 = (2b^2 - 13)^2 (b^2 - 5) * (b^2 - 5) + b^2 = (2b^2 - 13) * (2b^2 - 13) b^4 - 5b^2 - 5b^2 + 25 + b^2 = 4b^4 - 26b^2 - 26b^2 + 169 b^4 - 9b^2 + 25 = 4b^4 - 52b^2 + 169 Multiply by -1. That way, when the terms on the right side are moved to the left side, the b^4 coefficient will be positive. -b^4 + 9b^2 - 25 = -4b^4 + 52b^2 - 169 -b^4 + 4b^4 + 9b^2 - 52b^2 - 25 + 169 = 0 3b^4 - 43b^2 + 144 = 0 We need to factor the left side. If we use 12 and 12 as the factors for 144, we don't get the required coefficient for the b^2 term. The factors 16 and 9 will work because we multiply the 9 by 3. (3b^2 - 16)(b^2 - 9) = 0 Using those factors we have (3b^2 -16) = 0 or (b^2 - 9) = 0 Solving for b in the first factor: (3b^2 -16) = 0 3b^2 = 16 b^2 = 16/3 = 5.33333 b = 2.3094 This doesn't satisfy the requirement that all lengths are whole numbers. Solving for b in the second factor: (b^2 - 9) = 0 b^2 = 9 b = 3 Actually, b = -3 is also valid, but the length cannot be negative. We have side b = 3 miles, a = b^2 - 5 = 9 - 5 = 4 miles, c = 2b^2 - 13 = 2(9) - 13 = 18 - 13 = 5 miles So, we have a simple 3-4-5 right triangle.
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find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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distributive property

Since length times width equals area we can express the area of the entire field by adding the lengths of the two portions of the field and multiplying by the area as in the expression: 850(6y + 15x) = Area If we know that the area is 850000 square feet and we know that x = 50 feet we can rewrite the expression as: 850(6y + 15*50) = 943500 simplify to 850(6y + 750) = 943500 Now we have only one variable which we can solve for. Distrubute the 850 to by multiplying it by both terms inside of the parentheses to get 5100y + 637500 = 943500 subtract 637500 from both sides to get: 5100y = 306000 divide both sides by 5100 y = 60 Now substitute y back into the original equation: 850(60*6 + 750) = 943500 850(360 + 750) = 943500 By multiplying this we can see it checks out. 943500 = 943500 Now we need to find the area and length of each portion of the field So length of section 1 = 360 feet Length of section 2 = 750 feet   Area of section 1 = 850 x 360 = 306000 square feet Area of section 2 = 850 x 750 = 637500 square feet
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Two adjacent corrals are to be made using 240 ft of fencing. The fence must around the outer perimeter and across the middle. Find the dimensions so that the total enclosed area is as possible.

Two adjacent corrals are to be made using 240 ft of fencing. The fence must around the outer perimeter and across the middle. Find the dimensions so that the total enclosed area is as possible. The picture is a rectangle. The fwnce is across the middle of the rectangle. One side=y. The other side = x+x=2x. How do you know what variable to solve for first and how do you determine the domain? Thanks. You supplied no picture, but it's easy to draw one and solve the problem. Use two horizontal fences and three vertical fences, as shown in the picture I have below. As you can see, the area of one of the smaller corrals is x * y. By substituting the value of y in terms of x, we get x(80 - 4/3 x). Write this as an equation. y = x(80 - 4/3 x) y = 80x - 4/3 x^2 Set this equal to zero to simplify dealing with the left side of the equation. We will be working out the value of x where this graph crosses the x-axis, so y is naturally zero. 0 = 80x - 4/3 x^2 Divide by -4/3. 0 = -60x + x^2 x^2 - 60x = 0 Complete the square. Add (1/2 * -60)^2 to both sides of the equation. x^2 - 60x - 30^2 = 900   <<< that's -30 * -30 Factor the left side. (x - 30)(x - 30) = 900 Take the square root of both sides. x - 30 = ± 30 x = +30 + 30    and  x = - 30 + 30 x = 60          and  x = 0 This tells us that the graph of the equation is a parabola that crosses the x-axis at (0, 0) and (60, 0). The vertex, which is the highest point of the graph, is mid-way between those two points. The value of x at that point produces the maximum area for the smaller corral. Now, we calculate the value of y, using the formula derived in the picture. y = 80 - 4/3 x y = 80 - 4/3 (30) y = 80 - 4 (10) y = 80 - 40 y = 40 The two smaller corrals are 30 x 40; the large corral is 60 x 40 The corresponding areas are 1200 (for each of the two smaller corrals) and 2400 (for the large corral). The length of the fencing is 4 * 30 + 3 * 40 = 120 + 120 = 240. Now that we have the dimensions, we can see that the y sides are longer than the x sides, but the picture served its purpose.  
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Find the length of the boundary of the square field enclosing the lawn and the path

If ABCD is square, with side length x and a path of width y around the square, then the area of the path is: 4xy (the parts of the path bordering the square) 4y^2 (the square bits touching the corners of square ABCD). The area of the path is 4xy + 4y^2 Square ABCD, with an area of 729, has a side length of 27, so x = 27. The area of the path is given as 295, so: 4xy + 4y^2 = 295 4(27)y + 4y^2 = 295 108y + 4y^2 = 295 4y^2 + 108y - 295 = 0 Quadratic formula. . . y = (-108 +- sqrt( 108^2 - 4(4)(-295) ) ) / 2(4) y = (-108 +- sqrt(11664 + 4720) ) / 8 y = (-108 +- sqrt(16384) ) / 8 y = (-108 +- 128) / 8 It's a physical measurement, so it can't be negative. . . y = (-108 + 128) / 8 y = 20/8 y = 2.5 The width of the path is y = 2.5, but what about the outer square (with the path surrounding the inner square)? The outer square is x + 2y because the path sticks out on both sides of the inner square. 27 + 2(2.5) = 27 + 5 = 32 Answer:  The path is 2.5 meters wide, the inner square is 27 meters wide, and the outer square is 32 meters wide.
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how many square feet in an area 149.3' x 60.8' x 140' x 60'

Area has only two dimensions, whereas you have multiplied four dimensions! You say the area is irregular but you haven't defined the shape, so I'm going to assume you have a 4-sided area (quadrilateral) and you have provided the perimeter measurements. If we split the irregular quadrilateral into two triangles by joining the opposite corners, we may be able to use Heron's formula to find the area of the quadrilateral by adding the areas of the two triangles together. First we need to find the semiperimeter of the two triangles. We'll call the diagonal length x. Triangle 1 has a semiperimeter, S1, of 1/2(149.3+60.8+x) and triangle 2 a semiperimeter, S2, of 1/2(140+60+x). The area of triangle 1, A1=sqrt(S1(S1-149.3)(S1-60.8)(S1-x)) and A2=sqrt(S2(S2-140)(S2-60)(S2-x)). The area of the quadrilateral is A1+A2. S1=1/2(210.1+x)=105.05+x/2 and S2=1/2(200+x)=100+x/2. A1=sqrt((105.05+x/2)(x/2-44.25)(x/2+44.25)(105.05-x/2)) A2=sqrt((100+x/2)(x/2-40)(x/2+40)(100-x/2)) ​We now have to make another assumption, because we don't know what x is and the areas depend on it. The area is almost a rectangle or parallelogram, because 210 and 200 are similar measures, as are 60 and 60.8. Let's assume that triangle 2 is a right-angled triangle, then x=sqrt(140^2+60^2) (Pythagoras), so x=152.315 and x/2=76.158 (approx values). We can now calculate A1 and A2. A1=4484.92 and A2=4200, therefore the total area is 8684.92 sq ft. Note that A2 is half the area of a rectangle with sides 140 and 60=1/2(60*140)=4200. If we assume that triangle 1 is right-angled instead, what difference does it make? Its hypotenuse, x=sqrt(149.3^2+60.8^2)=161.21 and A1=4538.72 sq ft. If we calculate A2 using Heron's formula we get A2=4141.80, so total area is 8680.52 sq ft. So the difference between the two areas is less than 4.5 sq ft. We could average the two estimates of area: 8682.7 sq ft approx.
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