To determine which point lies on the graph of the equation (3x + 2y = 7x - 5), first, we can rewrite it in standard form. Rearranging gives (2y = 4x - 5) or (y = 2x - \frac{5}{2}). You can then test specific points (like integers) to see which satisfies this equation. For example, if you substitute (x = 2), then (y = 2(2) - \frac{5}{2} = 4 - 2.5 = 1.5), so the point (2, 1.5) lies on the graph.
The given expression is not an equation because it has no equality sign
if your point if 3/4 and your line is y=2x+4 then yes 3/4 does lie on this graph. any point in existence lies on this graph because its domain is all real numbers
It's the equation of a straight-line graph. Every point on the line is a solution of the equation. There are an infinite number of them.
To determine if the point (5, 6) is on the graph of the equation ( y - 2x + 5 = 0 ), we can substitute ( x = 5 ) into the equation. This gives us ( y - 2(5) + 5 = 0 ) or ( y - 10 + 5 = 0 ), which simplifies to ( y = 5 ). Since the point has coordinates (5, 6), and our calculation gives us ( y = 5 ), the point (5, 6) is not on the graph of the equation.
y = x + 3The graph of that equation is a straight line with a slope of 1 (45° rising to the right)and passing through the point y=3 on the y-axis.
The given expression is not an equation because it has no equality sign
if your point if 3/4 and your line is y=2x+4 then yes 3/4 does lie on this graph. any point in existence lies on this graph because its domain is all real numbers
(a) y = -3x + 1
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It's the equation of a straight-line graph. Every point on the line is a solution of the equation. There are an infinite number of them.
In this case, the discriminant is less than zero and the graph of this parabola lies above the x-axis. It never crosses.
To determine if the point (5, 6) is on the graph of the equation ( y - 2x + 5 = 0 ), we can substitute ( x = 5 ) into the equation. This gives us ( y - 2(5) + 5 = 0 ) or ( y - 10 + 5 = 0 ), which simplifies to ( y = 5 ). Since the point has coordinates (5, 6), and our calculation gives us ( y = 5 ), the point (5, 6) is not on the graph of the equation.
(5, -8)
On my graphing calculator, a TI84 Plus, I can enter the equation into the Y= (a button) and then graph it by hitting the Graph button.
Since no points were given, for any point (x,y), plug the x and y values of the point into the equation. If you get a contradiction, ie 5=3 or something similar, then the point does not lie on the graph.
(2, 11)
It rotates the graph about the point (0, b). The greater the value of m, the more steeply it rises to the right.