0
The range is just the difference between the largest and smallest values
lowest (0) and highest (6) for a range of 6.
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∙ 11y ago
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What is the slope of a line perpendicular to a line that contains the points 0 and 2 and -3 and 4?
Slope: (2-4)/(0--3) = -2/3 Perpendicular slope: 3/2
What is the standard form of the equation of a circle with its center at (2 -3) and passing through the point (-2 0)?
Points: (2, -3) and (-2, 0) Slope: -3/4 Equation: y = -0.75x-1.5
What is the distance between the points -2 0 and 5 3?
Let (x1, y1) = (-2, 0) and (x2, y2) = (5, 3). Distance between two points = square root of [(x2 - x1)2 + (y2 - y1)2] = square root of [(5 - -2)2 + (3 - 0)2] = square root of (72 + 32) = square root of 58
What ordered pairs is not a solution to the inequality y equals -3X - 2 A. 1. 1 B. 0. -1 C. 0. 0 D. -1. 0?
If y = -3x - 2 then substituting x from each ordered pair gives :- A) (1,1) y = (-3*1) - 2 = -5 ☒ B) (0,-1) y = (-3*0) - 2 = -2 ☒ C) (0,0) y = (-3*0) - 2 = -2 ☒ D) (-1,0) y = (-3*-1) - 2 = 1 ☒ So the answer is ALL OF THEM are not solutions to the equation y = -3x - 2.......BUT, you've used the word Inequality so depending whether y > -3x - 2 or y < -3x -2 clearly affects the results.
How do you work out the solutions for 3x-2y equals 1 and 3x square-2y square plus 5 equals 0?
How to solve: 3x - 2y = 1 3x2 - 2y2 + 5 = 0 Rearrange the first equation to make x or y the subject (that is x = something or y = something) and then substitute into the second equation and solve that: 3x - 2y = 1 => y = (3x - 1)/2 3x2 - 2y2 + 5 = 0 => 3x2 - 2((3x - 1)/2)2 + 5 = 0 [substitute for y] => 3x2 - 2(9x2 - 6x + 1)/4 + 5 = 0 [expand the square term] => 3x2 - (9x2 - 6x + 1)/2 + 5 = 0 [spot that 2w/4 is the same as w/2] => 6x2 - (9x2 - 6x + 1) + 10 = 0 [multiply equation by 2] => 6x2 - 9x2 + 6x - 1 + 10 = 0 [remove the brackets by multiplying by -1 as it is -1 x (..)] => -3x2 + 6x + 9 = 0 [collect together terms] => 3x2 - 6x - 9 = 0 [multiply whole equation by -1] => x2 - 2x - 3 = 0 [divide whole equation by 3] => (x - 3)(x + 1) =0 [factorize) => x = 3 or -1 [as one factor or the other must be zero] Now use first equation to find corresponding y terms: x = 3:y = (3 x (3) - 1) / 2 = 8 / 2 = 4 x = -1: y= (3 x (-1) - 1) /2 = -4 / 2 = -2 So the solution is the (x, y) pairs, or points, (3, 4) and (-1, -2). The answer can be checked using the second equation: (3, 4): 3(3)2 - 2(4)2 + 5 = 3 x 9 - 2 x 16 + 5 = 27 - 32 + 5 = 0 (-1, -2): 3(-1)2 - 2(-2)2 + 5 = 3 x 1 - 2 x 4 + 5 = 3 - 8 + 5 = 0
What is the range of this relation (2 -3) (-4 2) (6 2) (-5 -3) (-3 0)?
The range is {-3, 2, 0}.
What is the range of this relation(2, -3), (-4, 2), (6, 2), (-5, -3), (-3, 0)?
{-3,2,0}
What is the range of this relation(-3, 2), (2, -4), (2, 6), (-3, -5), (0, -3)?
{2,-4,6,-5,-3}
What is the range of this relation -3 2 2 -4 2 6 -3 -5 0 -3?
1 it is the difference between the highest and lowest number
What is the range of 3 3 4 2 1 3 3?
3,3,4,2,1,3,3 3-3=0 range=0
Is this relation a function{(0, 0), (1, 0), (2, 0), (3, 0), (4, 0)}?
yes
What is the domain of this relation(-3, 2), ( 2, -4), (2, 6), (-3, -5), (0, -3)?
{-3,2,0}
What is the range of data with numbers 3 0 3 4 1 6 2 4 3 2 0 3 5?
The range is 6. (6 - 0 = 6)
What is the domain of this relation(2, -3), (-4, 2), (6, 2), (-5, -3), (-3, 0)?
{2,-4,6,-5,-3}
What is the range of 0 1 2 3 4 5 6?
The range of {0, 1, 2, 3, 4, 5, 6} is 6 - 0 = 6.
Is the relation (1 3) (4 0) (3 1) (0 4) (2 3) a function?
If those are the only values, no.
What is the range and mean of 0 2 2 2 3 4 8?
Mean: 3 I think the range is 7 or 8
