Let the required equation be y = mx + c; where y = 1, m = f'(0), x = 0
f(a) = sec(a)
f'(a) = sec(a)tan(a)
f'(0) = sec(0)tan(0) = 0
y = mx + c
1 = 0(0) + c
c = 1
Therefore, required equation is
y = 1.
Split it into 7 triangles. Half-base of each is 2 so the distance of center to the midpoint of the base is (Pythagoras) square root of (4.62 - 22) = 4.162 triangle is this times 2, and with 7 triangles multiply also by 7 to get an approximate area of 58 sq in.
As far as the old dimensions of the photo are concerned,
Width of the photo = 3"
Length of the photo = 5"
New length of the photo = 7 inches
Let us assume the new width of the photo = x inches
Then
3/5 = x/7
5x = 7 * 3
5x = 21
x = 21/5
= 4.20 inches
So the correct option among all the options that are given in the question is the first option. I hope that the answer has come to your help.
