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The browser used by this site for posting questions is all but useless for mathematics because it rejects most symbols. It is, therefore, not clear what your question is. If it was 2x = y - 1 then the answer consists of every point on the line which is defined by the equation.
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What is the answer to 3x y1 x y1?
{3x +y =1 {x+y= -3
What does the point slope formula look like?
y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is a point on the line.
How do you write an equation given two points?
If there are given two points, (x1, y1) and (x2, y2), then you can write the equation of a line by finding the slope first [slope = m = (y2 - y1)/(x2 - x1)] and using one of the points in order to write the equation in the point-slope form such as(y - y1) = m(x - x1)y - y1 = mx - mx1y = mx - mx1 + y1y = mx + (y1 - mx1) the slope-intercept form, where m is the slope and (y1 - mx1) is the y-intercept.mx - y = mx1 - y1 the general form of the equation of the line.
What is a point slope form equation?
(y-y1)=m(x-x1) OR we can write it y=m(x-x1)+y1
What formula is used to find the length between two points in a coordinate plane?
It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]
How do you make a heart on a graphing calculator?
Y1=2x^(2/3)+√(20-x²)-3 Y2=2x^(2/3)-√(20-x²)-3
What is the equation for grphing a heart on a calculator?
Here it is: Y1=2x^(2/3)+√(20-x²)-3 Y2=2x^(2/3)-√(20-x²)-3
What is the inverse of a function y1?
Definition of the inverse of a function.Let f and g be two functions such thatf(g(x)) = x for every x in the domain of g andg(f(x)) = x for every x in the domain of f.The function g is the inverse of the function f, and the domain of f is equal to the range of g, and vice versa.Example: Find the inverse of y1 = 2x + 7Solutiony1 = 2x + 7 interchange x and y;x = 2y1 + 7 solve for y;x - 7 = 2y1 + 7 -7 subtract 7 to both sides;x - 7 = 2y1 divide by 2 both sides;(x - 7)/2 = y1 replace y1 with y2;y2 = (x - 7)/2Thus, the inverse of y1 = 2x +7 is y2 = (x -7)/2Let's check if this is true according to the above definition:Let y1 = f(x) = 2x +7 and y2 = g(x) = (x -7)/21. f(g(x))= x ?f(x) = 2x + 7f((x - 7)/2) = 2[(x -7)/2] + 7 = x - 7 + 7 = x True2. g(f(x) = x ?g(x) = (x - 7)/2g(2x + 7) = [(2x + 7) - 7]/2 = 2x/2 = x True
If f is a differentiable function such that f of 9 equals 2 and f' of 9 equals -2 then what is the linearization?
Use the point-slope form: y - y1 = m(x - x1). From the givens, y1 = 2, x1 = 9, and m = -2. Thus, y - 2 = -2(x - 9) = -2x + 18, or y = -2x + 20.
What is the equation of the line that goes through -1 3 and 5 -9?
(-1,3),(5,-9) (y2-y1)/(x2-x1)= (-9-3)/(5--1)= (-12/6)= the slope is -2 m(x-x1)=y-y1 -2(x-5)=y--9 -2x+10=y+9 y=-2x+1
Write the equation in slope intercept form of the line that has a slope of 2 and contains the point 3 7?
y-y1=m(x-x1) y-7=2(x-3) y-7=2x-6 y=2x-1 y=2x-1
Write the equation in slope-intercept form of the line that has a slope of 2 and contains the point 3 7?
Use the slope intercept form here, Y - Y1 = m(X - X1) m = 2 Y1 and X1 = (3, 7) (Y - 7) = 2(X - 3) Y - 7 = 2X - 6 Y = 2X + 1 ----------------the equation of the line
What formula with functions can you enter to calculate the range of values in Y1 to Y10?
There are many calculations that could be done: =SUM(Y1:Y10) =AVERAGE(Y1:Y10) =MAX(Y1:Y10) =MIN(Y1:Y10) =COUNT(Y1:Y10)
What is 5x-y1?
5x-y1 = 4
What is the equation of the curve yx2 plus 2x plus 1 when y1?
Without an equality sign the given terms of an expression can't be considered to be an equation and so therefore no solution is possible.
What is th eqaution of the line tha passes through 3 -5 and 2 -3?
To find the line that passes through the points (3, -5) and (2, -3), you can plug those numbers into the classic equation for a line: Δy = sΔx or: (y - y1) = s(x - x1) You can calculate the slope given the numbers that are already provided: s = Δy/Δx s = ((-5) - (-3)) / (3 - 2) s = (3 - 5) / (3 - 2) s = -2/1 s = -2 Now that you have the slope, you can use it in the original line equation, plugging either of the two points in to your x1 and y1 values: (y - y1) = s(x - x1) y - (-5) = -2(x - 3) y + 5 = -2x + 6 y = 1 - 2x You could also plug in the other pair of x/y values to get the same result: (y - y1) = s(x - x1) y - (-3) = -2(x - 2) y + 3 = -2x + 4 y = 1 - 2x
What is the y intercept of this line 2x y1?
Unfortunately, limitations of the browser that is used by WA means that most symbols cannot be seen. Please resubmit using words like "plus", "minus" and "equals".
