The value of the angle is 105 degrees.
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:
aka 9 units
Step-by-step explanation:
Its pretty simple so what u do is count the sqaures then multiply them and ill do it for you
H=3
B=6
and the formula for a triangle is A=
so its 3x6/2=9
The answer is 9
<span>decimal 1.25
percent 125%
fraction 5/4</span>
first you find the slope by making using the quation for slope
(7-3)/(0-5) and you find your slope is 4/-5. Then you can plug this into point slope form using on of the coordinates so:
y-y1=m(x-x1) --> y-3=-4/5(x-5)
then you distribute the -1 so -----> y-3=-4/5x+4 and then move the 3 over
so the answer is y= -4/5x+7
