No.
Since x + (x-1) + 2x = 180 degrees
ie 4x - 1 = 180 degrees, if follows that 2x > 90 degrees.
So the largest angle is obtuse.
it divides theline segment..xDusing midpoint formula..and division of line segment formula..m=(X1+X2)/2 (Y1+Y2)/2X=X1+r(X2-X1)xD ..
(-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
To find an edge of a triangle, you can use the distance formula if you have the coordinates of the triangle's vertices. For vertices A(x1, y1) and B(x2, y2), the length of edge AB can be calculated using the formula: ( \text{Distance} = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} ). Alternatively, if you know the triangle's angles and one side length, you can apply the Law of Sines or Law of Cosines to find the lengths of other edges.
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
1.) Triangle method 2.)Formula- Y2-Y1 -------- X2-X1 3.)two points
x²+2x-15=0 x1=-2/2 - Square root of ((2/2)²+15) x1=-1-4 x1=-5 x2=-2/2 + Square root of ((2/2)²+15) x2=-1+4 x2=3
In QBasic, you can color a triangle using the LINE statement to draw the triangle's sides and the PAINT statement to fill it with color. First, use LINE to specify the coordinates of the triangle's vertices. Then, use PAINT at one of the triangle's vertices with the desired fill color. For example: LINE (x1, y1)-(x2, y2), color LINE (x2, y2)-(x3, y3), color LINE (x3, y3)-(x1, y1), color PAINT (x1, y1), fillColor Replace (x1, y1), (x2, y2), (x3, y3) with the triangle's coordinates and color and fillColor with your chosen colors.
it divides theline segment..xDusing midpoint formula..and division of line segment formula..m=(X1+X2)/2 (Y1+Y2)/2X=X1+r(X2-X1)xD ..
The degree of the polynomial 2x + 5 is 1. The highest power of x is x1, i.e. 2x1 + 5x0, hence the designation of first degree.
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.
(-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
To find an edge of a triangle, you can use the distance formula if you have the coordinates of the triangle's vertices. For vertices A(x1, y1) and B(x2, y2), the length of edge AB can be calculated using the formula: ( \text{Distance} = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} ). Alternatively, if you know the triangle's angles and one side length, you can apply the Law of Sines or Law of Cosines to find the lengths of other edges.
To factor this, write the equation 2x2 - 2x - 2 = 0, and solve with the quadratic equation. You may get two numbers, which we may call "x1" and "x2"; in this case, the factorization is (x - x1)(x - x2).
Z = 3x1+5x2+4x3 Subject to constraints2 x1 + 3x2 =8 3x1 + 2x 2 + 4x3=15 2x2 + 5x3 = 10 x1,x2,x3, =0
y-y1=m(x-x1) y-7=2(x-3) y-7=2x-6 y=2x-1 y=2x-1
the questions is 2x=(cot^2 x-1)/(cot^2 x+1)
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