Y = e^tanx - 2
To find at which point it crosses x axis we state that y= 0
e^tanx - 2 = 0
e^tanx = 2
tanx = ln 2
tanx = 0.69314
x = 0.6061
to find slope at that point first we need to find first derivative of funtion y.
y' = (e^tanx)*1/cos^2(x)
now we express x = 0.6061 in y' and we get:
y' = k = 2,9599
Answer:
-24+12(d-3)+22=-24+34(d-3)
10(d-3)
10d=-30
d=-30/10
d=3
Answer:
Step-by-step explanation:
36. (4,1)
x-axis(4,-1)
y-axis(-4,1)
37.(-2,3)
x-axis(2,-3)
y-axis(-2,3)
38.(2,-5)
x-axis(-2,5)
y-axis(2,-5)
39.(-3.5, -2.5)
x-axis(-3.5,2.5)
y-axis(3.5,-2.5)
Please correct me I am wrong
Here is the y-axis formula (-x,y)
Here is the x-axis formula(x,-y)