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What else can I help you with?
How do you find slope if the points aren't given?
the slope is the 'm' in y=mx+b so even if the points aren't given, if there is an equation, then you can find the slope. for example, if you have an equation like this: y=2x+5 the slope is 2 and the y-intercept is 5.
Find an equation of the line containing the given point and having the given slope 3-6m1?
Assuming the point is (3, -6) and the slope 1, the equation is x - y - 9 = 0
How do you know if a quadratic equation will have one two or no solutions How do you find a quadratic equation if you are only given the solution Is it possible to have different quadratic equation?
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
When two planes equation given and one parametric equation is given how can you find the plane equation?
If you have two equations give AND one parametric equation why do you need to find yet another equation?
How do you find The equation of a line given two points needed?
First, you calculate the slope between the two points (difference of y / difference of x). Then you can use the equation, using one of the points (x1, y1): y - y1 = m(x - x1) Just replace x1 and y1 with the coordinates of the point, and m with with the slope.
Find the equation of the line containing the two points shown?
what is the slope of the line containing points (5-,-2) and (-5,3)? 2
How do you find slope if the points aren't given?
the slope is the 'm' in y=mx+b so even if the points aren't given, if there is an equation, then you can find the slope. for example, if you have an equation like this: y=2x+5 the slope is 2 and the y-intercept is 5.
How do you find Slope intercept equation of given two points?
Use the equation; y=mx+b where m is the slope Use your 2 points as y and b (intercept)
Find an equation of the line containing the given point and having the given slope 3-6m1?
Assuming the point is (3, -6) and the slope 1, the equation is x - y - 9 = 0
Can you find an equation of the line containing the given pair of points 0 -6 and 4 2?
Suppose the equation of the line is y = mx + c where m and c need to be determined. The slope of the line = (difference in y-coordinates of the two given points)/(difference in x-coordinates of the two given points) = (-6 - 2)/(0 - 4) = -8/-4 = 2 So m = 2 ie the equation of the line becomes: y = 2x + c where c still needs to be determined. The point (0, -6) is on the line. That is, when x = 0, y = -6. Substituting in the equation, -6 = 2*0 + c so that c = -6 and the equation of the line is y = 2x - 6
Find an equation of the line having the given slope and containing the given point m equals 2 over 3 and 2 and -1?
If the slope is 2/3 and the coordinate is (2, -1) then the straight line equation is 3y=2x-7
How do you know if a quadratic equation will have one two or no solutions How do you find a quadratic equation if you are only given the solution Is it possible to have different quadratic equation?
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
When two planes equation given and one parametric equation is given how can you find the plane equation?
If you have two equations give AND one parametric equation why do you need to find yet another equation?
How do you find the equation of a parabola in standard form given 3 points?
I suggest that the simplest way is as follows:Assume the equation is of the form y = ax2 + bx + c.Substitute the coordinates of the three points to obtain three equations in a, b and c.Solve these three equations to find the values of a, b and c.
How do you find the value of y in algebra?
You solve an equation containing y.
How do you find the equation of a tangent line?
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
Is it easier to solve an equation for y first before finding points?
It is always easier to use an equation to find points since all you would have to do is substitute values into the equation to find the final unknown value that will tell the point. To get the equation, however, you would usually need to have some points at the start to help derive the equation in the end.
