Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio
n. Do not attempt to find the solution. (Enter your answer using interval notation.) (x − 2)y'' + y' + (x − 2)(tan x)y = 0, y(3) = 1, y'(3) = 4
1 answer:
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Answer:
Length of the rectangle is 15.0325 km and width is 0.3765 km.
Explanation:
Given:
Perimeter of a rectangle = 30.8 km
Length of diagonal of rectangle = 11 km
To find:
The length and width of rectangle=?
Solution:
Lets assume length of the rectangle = x km
And assume width of the rectangle = y km
Lets first create equation using given perimeter
perimeter of rectangle = 2 ( length + width )
=> 30.8 km = 2 ( x + y )
=>
=> y = 15.4 – x ------(1)
As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,
=>
=>
=>
On substituting value of y from (1) in above equation we get
=>
=>
=>
=>
Solving above quadratic equation using quadratic formula
General form of quadratic equation is
And quadratic formula for getting roots of quadratic equation is
As equation is , in our case
a = 2 , b = -30.8 and c = 116.16
Calculating roots of the equation we get
=> x = 15.0325 or x = 0.3675
As generally length is longer one ,
So x = 1.0325
From equation (1) y = 15.4 – x = 0.3765
Hence length of the rectangle is 15.0325 km and width is 0.3765 km.