0.025 is less than 0.052.
Let us find the co ordinates of each vertex of the triangle .
Vertex A ( in firs second quadrant) = ( -5 ,3)
vertex B in third quadrant = ( -5, -5)
vertex C in fourth quadrant = ( 4, -2)
let us use distance formula AB^2 = ( -5 - 3)^2 + (-5 - -5 )^2 = 64 + 0
AB= 8
BC^2 = ( -2 - -5 )^2 + ( 4 - - 5)^2 = 9 + 81 = 90
BC = 9.48
AC^2 = ( -2 -3)^2 +( 4- -5)^2 = 25 + 81 = 106
AC= 10.29
Perimeter = sum of length of AB+ BC+ Ac = 8 + 10.29 + 9.48= 22.77
Given one to one f(6)=3, f'(6)=5, find tangent line to y=f⁻¹(x) at (3,6)
OK, we know
f⁻¹(3)=6
The inverse function is the reflection in y=x. So slopes, i.e. the derivative will be the reciprocal. We know the derivative of f at 6 is 5, so the derivative of f⁻¹ at y=6 is 1/5, which corresponds to x=3.
f⁻¹ ' (3) = 1/5
That slope through (3,6) is the tangent line we seek:
y - 6 = (1/5) (x-3)
That's the tangent line.
y = x/5 + 27/5
Answer: 0.88888889
Step-by-step explanation:
Just divide 8/9!