The figure below shows a partially completed set of steps to construct a rhombus PQRS:The next step is to construct side PS of the figure so that side PS is parallel and congruent to side QR. The chart below shows the steps each of the four students took to draw side PS:Student 1 Fix the compass at M and adjust its width to point L. Without changing the width of the compass, move the compass to N and draw a small arc on the big arc. Label the point of intersection of the two arcs as T. Draw a line segment from P that passes through T. Adjust the width of the compass to QR and draw an arc from point P to intersect line PT at S.Student 2 Fix the compass at M and draw an arc that intersects side QP at point T. Without changing the width of the compass, move the compass to R and draw an arc. Adjust the width of the compass to QR and draw an arc from point P. Draw a line segment from R that passes through T and intersects the second arc at S.Student 3 Fix the compass at L and adjust its width to point M. Without changing the width of the compass, move the compass to R and draw an arc which intersects QR at point T. Draw a line segment from R that passes through T and intersects the arc at S.Student 4 Fix the compass at L and adjust its width to point P. Without changing the width of the compass, move the compass to R and draw an arc which intersects QR at point T. Draw a line segment from R that passes through T and intersects the arc at S.Which student used the correct steps to construct rhombus PQRS?A. Student 2B. Student 4C. Student 3D. Student 1

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general 11 months ago 8873

PLEASE HELP8.011. Find the vertex, focus, directrix, and focal width of the parabola.x2 = 20y A) Vertex: (0, 0); Focus: (0, 5); Directrix: y = -5; Focal width: 20 B) Vertex: (0, 0); Focus: (5, 0); Directrix: x = 5; Focal width: 5 C) Vertex: (0, 0); Focus: (5, 0); Directrix: y = 5; Focal width: 80 D) Vertex: (0, 0); Focus: (0, -5); Directrix: x = -5; Focal width: 802. Find the vertex, focus, directrix, and focal width of the parabola.x = 3y2 A) Vertex: (0, 0); Focus: one divided by twelve comma zero ; Directrix: x = one divided by twelve ; Focal width: 12 B) Vertex: (0, 0); Focus: the point one twelfth comma zero ; Directrix: x = negative one twelfth ; Focal width: 0.33 C) Vertex: (0, 0); Focus: zero comma one divided by sixteen ; Directrix: x = negative one divided by sixteen ; Focal width: 0.33 D) Vertex: (0, 0); Focus: one divided by sixteen comma zero ; Directrix: y = nnegative one divided by sixteen ; Focal width: 123. Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, 9). (1 point) A) y = one divided by thirty six x2 B) y = one divided by nine x2 C) y2 = 9x D) y2 = 36x4. Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8. A) y = one divided by thirty two x2 B) y2 = 8x C) y2 = 32x D) y = one divided by eight x25. Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7. A) y = one divided by twenty eight x2 B) x = one divided by twenty eight y2 C) -28y = x2 D) y2 = 14x6. A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base, as shown below.A parabola opening down with vertex at the origin is graphed on the coordinate plane. The height of the parabola from top to bottom is seventy four feet and its width from left to right is twenty eight feet.Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

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general 11 months ago 4549

PLEASE HELP8.021. Find the center, vertices, and foci of the ellipse with equation x squared divided by one hundred plus y squared divided by sixty four = 1. A) Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-8, 0), (8, 0) B) Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -8), (0, 8) C) Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-6, 0), (6, 0) D) Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -6), (0, 6)2. Find the center, vertices, and foci of the ellipse with equation 2x2 + 6y2 = 12. A) Center: (0, 0); Vertices: point negative square root of six comma zero and point square root of six comma zero ; Foci: (-2, 0), (2, 0) B) Center: (0, 0); Vertices: (0, -6), (0, 6); Foci: Ordered pair zero comma negative four square root two and ordered pair zero comma four square root two ; C) Center: (0, 0); Vertices: point zero comma negative square root of six and point zero comma square root of six ; Foci: (0, -2), (0, 2) D) Center: (0, 0); Vertices: (-6, 0), (6, 0); Foci: Ordered pair negative four square root two comma zero and ordered pair four square root two comma zero3. Graph the ellipse with equation x squared divided by twenty five plus y squared divided by four = 1. A) A vertical ellipse is shown on the coordinate plane centered at, five, two, with vertices at, five, seven, and five, negative three and minor axis endpoints at three, two, and seven, two. B) A horizontal ellipse is shown on the coordinate plane centered at five, two, with vertices at zero, two, and ten, two and minor axis endpoints at five, four and five, zero. C) A vertical ellipse is shown on the coordinate plane centered at the origin with vertices at zero, five and zero, negative five and minor axis endpoints at, negative two, zero and two, zero. D) A horizontal ellipse is shown on the coordinate plane centered at the origin with vertices at, negative five, zero and, five, zero and minor axis endpoints at, zero, two and zero, negative two.4. Find an equation in standard form for the ellipse with the vertical major axis of length 6 and minor axis of length 4. A) x squared divided by two plus y squared divided by three = 1 B) x squared divided by nine plus y squared divided by four = 1 C) x squared divided by three plus y squared divided by two = 1 D) x squared divided by four plus y squared divided by nine = 15. An elliptical riding path is to be built on a rectangular piece of property, as shown below.A vertical ellipse is shown centered on the coordinate plane, surrounded by a rectangle of equal length and width.The rectangular piece of property measures 8 mi by 6 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides.

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general 11 months ago 5234

PLEASE HELP8.03a1. Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1. A) Vertices: (± 5, 0); Foci: (± 3, 0) B) Vertices: (± 3, 0); Foci: (± 5, 0) C) Vertices: (0, ± 3); Foci: (0, ±5) D) Vertices: (0, ± 5); Foci: (0, ± 3)2. Find the vertices and foci of the hyperbola with equation quantity x minus three squared divided by sixteen minus the quantity of y plus four squared divided by nine = 1. A) Vertices: (7, -4), (-1, -4); Foci: (-2, -4), (8, -4) B) Vertices: (-4, 7), (-4, -1); Foci: (-4, -2), (-4, 8) C) Vertices: (6, -4), (0, -4); Foci: (0, -4), (6, -4) D) Vertices: (-4, 6), (-4, 0); Foci: (-4, 0), (-4, 6)3. Graph the hyperbola with equation A horizontal hyperbola is shown on the coordinate plane centered at, negative one, four, with vertices at, negative four, negative four and two, negative four. = 1. A) A vertical hyperbola is shown on the coordinate plane centered at the origin with vertices at, zero, two and zero, negative two. B) A horizontal hyperbola is shown on the coordinate plane centered at, negative three, three, with vertices at, negative five, three and negative 1, three. C) A vertical hyperbola is shown on the coordinate plane centered at, three, negative three, with vertices at, three, negative one and three, negative five. D) A horizontal hyperbola is shown on the coordinate plane centered at the origin with vertices at, negative two, zero and two, zero.4. Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7). A) y squared over forty nine minus x squared over four = 1 B) y squared over forty five minus x squared over four = 1 C) y squared over four minus x squared over forty nine = 1 D) y squared over four minus x squared over forty five = 15. Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ± three divided by four times x..

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general 11 months ago 3935