Let . The tangent plane to the surface at (0, 0, 8) is
The gradient is
so the tangent plane's equation is
The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation
or , , and .
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
Answer:
point - slope form
y - 1 = 3 (x-0)
Step-by-step explanation:
<u><em>step(i):-</em></u>
Given points are (0,1) and (2,7)
The slope of the line
m =3
<u><em>Step(ii):-</em></u>
point -slope form
y - y₁ = m(x-x₁)
Equation of the straight line passing through the point (0,1) and having slope 'm' = 3
y - 1 = 3 (x-0)
y -1 = 3x
3x - y +1=0
Equation of the straight line passing through the point (0,1) and having slope 'm' = 3 is 3x -y +1=0
62.5/100 is the batting average as a fraction
Answer:
y = 0.5x
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = 0.5x - 2 ← is in slope- intercept form
with slope m = 0.5
Parallel lines have equal slopes, thus
y = 0.5x + c ← is the partial equation
To find c substitute (- 2, 1) into the partial equation
- 1 = - 1 + c ⇒ - 1 + 1 = 0
y = 0.5x + 0 , that is
y = 0.5x ← equation of parallel line