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graph the points on the given coordinate plane. then name the quadrant …

Question

graph the points on the given coordinate plane. then name the quadrant or axis on which each point lies.

  1. s(3,0)
  2. r(4,1)
  3. c(3,−3)
  4. t(2,1)
  5. u(−3,−3)
  6. v(−3,3)
  7. w(3,4)
  8. d(1,−3)
  9. a(−4,−2)
  10. b(−2,0)
  11. q(0,3)
  12. z(−1,−4)

identify the quadrant or axis that indicates the location of each point.

  1. (7,−8)
  2. (−10,10)
  3. (−14,0)
  4. (−11,−16)
  5. (18,0)
  6. (9,13)
  7. (0,8)
  8. (14,5)
  9. (9,−4)
  10. (−12,−5)
  11. (23,0)
  12. (−1,8)

solve. show your work.

  1. if the points a(−5,4), b(4,4); c(4,−4); and d(−5,−4) were graphed on a coordinate plane and connected, what figure would be created? name one point in each quadrant and on each axis that is inside this figure.
  2. the teacher purchased 5 cases of p

within each case there were 4 box
pencils that each contained one do
pencils. there are 30 students in t
if the teacher gives each student\tnumber of pencils, how many ca
student receive? are there any p
over? if so, how many?

Explanation:

Response
Problem 49:

Step1: Analyze coordinates of points

Point \( A(-5, 4) \): \( x=-5 \) (negative), \( y = 4 \) (positive) – Quadrant II.
Point \( B(4, 4) \): \( x = 4 \) (positive), \( y = 4 \) (positive) – Quadrant I.
Point \( C(4, -4) \): \( x = 4 \) (positive), \( y=-4 \) (negative) – Quadrant IV.
Point \( D(-5, -4) \): \( x=-5 \) (negative), \( y=-4 \) (negative) – Quadrant III.

Step2: Check side lengths and angles

Distance \( AB \): \( \vert 4 - (-5)\vert=\vert9\vert = 9 \) (horizontal, same \( y \)-coordinate).
Distance \( BC \): \( \vert -4 - 4\vert=\vert -8\vert = 8 \) (vertical, same \( x \)-coordinate).
Distance \( CD \): \( \vert -5 - 4\vert=\vert -9\vert = 9 \) (horizontal, same \( y \)-coordinate).
Distance \( DA \): \( \vert 4 - (-4)\vert=\vert 8\vert = 8 \) (vertical, same \( x \)-coordinate).
All angles between sides are \( 90^\circ \) (since horizontal and vertical lines are perpendicular).

Step3: Identify the figure

A quadrilateral with four right angles and opposite sides equal is a rectangle. Also, since \( AB = CD = 9 \) and \( BC = DA = 8 \), and all angles are right angles, it is a rectangle (specifically, a rectangle with length 9 and width 8; it can also be considered a rectangle, not a square as length ≠ width).

Step4: Find points inside

  • Quadrant I: \( (1, 1) \) (inside, \( x>0, y>0 \), within \( x \in [-5,4] \), \( y \in [-4,4] \)).
  • Quadrant II: \( (-1, 1) \) (inside, \( x<0, y>0 \)).
  • Quadrant III: \( (-1, -1) \) (inside, \( x<0, y<0 \)).
  • Quadrant IV: \( (1, -1) \) (inside, \( x>0, y<0 \)).
  • On x - axis: \( (0, 0) \) (inside, \( y = 0 \), \( x \in [-5,4] \)).
  • On y - axis: \( (0, 0) \) (inside, \( x = 0 \), \( y \in [-4,4] \)) or \( (0, 1) \), etc.

Step1: Calculate total pencils

  • Cases: 5, Boxes per case: 4, Dozen pencils per box: 12 (since 1 dozen = 12).

Total pencils \( = 5\times4\times12 \).
\( 5\times4 = 20 \), \( 20\times12 = 240 \) pencils.

Step2: Distribute to students

Students: 30. Let \( x \) be pencils per student.
\( 30x\leq240 \), \( x=\frac{240}{30}=8 \).

Step3: Check remainder

\( 240 - 30\times8 = 240 - 240 = 0 \).

Answer:

The figure is a rectangle.

  • Quadrant I: \( (1, 1) \)
  • Quadrant II: \( (-1, 1) \)
  • Quadrant III: \( (-1, -1) \)
  • Quadrant IV: \( (1, -1) \)
  • On x - axis: \( (0, 0) \)
  • On y - axis: \( (0, 0) \)
Problem 50 (Assuming missing words: "pencils", "boxes", "dozen", "the same"):