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What are three different ordered pairs that are solutions of the equation Y equals x 5?
In ordered pair forms (0,-5) (1,-4) (2,-3) The first values are the x values and the second values are the y values.
X plus y 6?
I assume you meant, x + y = 6.Now you can get multiple answers (if x and y are integers).x = 0, y = 6x = 1, y = 5x = 2, y = 4x = 3, y = 3x = 4, y = 2x = 5, y = 1x = 6, y = 0
What ordered pairs satisfy the equation y equals 1 5x?
To find the ordered pairs in any equation, just plug in any number for x and solve for y. If your equation is meant to be y=1+5x, then if x=0 then y=1+5*0, y=1 so the first ordered pair would be (0,1) If your equation is meant to be y=(1/5)x, then if x=0 then y=(1/5)*0, y=0, so the first ordered pair would be (0,0)
What is the radius of this circle (x - 3)2 plus (y plus 1)24?
If its meant to be: (x-3)^2+(y+1)^2 = 4 then the radius is 2
What is the value of two numbers if their sum is 24 and their difference is 4?
We let the two numbers be represented by x and y. Then x+y=24 x-y=4 Adding equations yields 2x=28 Or x=14, which forces y=10. 10+14=24, 14-10=4.
What are the factors of 45x4y?
If you meant (45x^4)(y), then the answer is: (3)(3)(5)(x)(x)(x)(x)(y) If you meant 45 times x times 4 times y, then the answer is: (2)(2)(3)(3)(5)(x)(y)
Points of y equals -2x plus 6 and x equals 1?
Given: y = -2x + 6 x = 1 y = -2(1) + 6 y = 4 I believe this is what you meant. Possibly you meant: y = -2x + 6 + x = 1 Then you would get y = 1 -x + 6 = 1 x = 5
What are three different ordered pairs that are solutions of the equation Y equals x 5?
In ordered pair forms (0,-5) (1,-4) (2,-3) The first values are the x values and the second values are the y values.
What is the x-intercept of the line given by the equation x- 4y20?
Assuming that you meant "x - 4y = 20", the x-intercept is x=20, and the y-intercept is y=-5.
How do you graph x plus y equals 5 and x plus y equals 6?
you subtract x from both sides for both equations to get it in y= form. so... x+y=5 x+y-x=5-x y=5-x x+y=6 x+y-x=6-x y=6-x thanks:D but i meant with absolute value signs it was supposed to look like this : |x+y|=5 and |x|+|y|=6 ... but when i typed it in answers.com changed it
X plus y plus z squared?
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
How do you solve the net force?
Usually you would add individual forces. You have to add them as vectors. You can do this graphically, or by adding the components (x, y, z) separately.Usually you would add individual forces. You have to add them as vectors. You can do this graphically, or by adding the components (x, y, z) separately.Usually you would add individual forces. You have to add them as vectors. You can do this graphically, or by adding the components (x, y, z) separately.Usually you would add individual forces. You have to add them as vectors. You can do this graphically, or by adding the components (x, y, z) separately.
The sum of two forces acting at a point is 16 Newton If the resultant force is 8 Newton and its direction is perpendicular to minimum force then the forces are?
The two forces are 8 Newton each, acting in perpendicular directions. The minimum force creates a right triangle with the resultant force, such that the square of the minimum force plus the square of 8 Newton force equals the square of 16 Newton (by using Pythagoras theorem).
X plus y 6?
I assume you meant, x + y = 6.Now you can get multiple answers (if x and y are integers).x = 0, y = 6x = 1, y = 5x = 2, y = 4x = 3, y = 3x = 4, y = 2x = 5, y = 1x = 6, y = 0
What is meant when you say dependent variable y does not depend on independent variable x?
you can change the value of x to any new value and it has no effect on the value of y.
How do you find the x and y intercept of -4 plus 8y equals 16?
Your equation does not have an x variable. So as-is would be a horizontal line (no x-intercept). If this was a typo and you meant to have x in one of the terms then it would have an x and y intercept. The y intercept is where it intersects the y-axis. The x coordinate of the y-axis is x=0, so substitute x=0 into the equation, and solve for y. To find the x-intercept, substitute y=0, and solve for x.
What ordered pairs satisfy the equation y equals 1 5x?
To find the ordered pairs in any equation, just plug in any number for x and solve for y. If your equation is meant to be y=1+5x, then if x=0 then y=1+5*0, y=1 so the first ordered pair would be (0,1) If your equation is meant to be y=(1/5)x, then if x=0 then y=(1/5)*0, y=0, so the first ordered pair would be (0,0)
