What is the slope of a line perpendicular to the line with equation y = 1/6x – 2? help pls very urgent
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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.
a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
b. The inverse image of g(x) is
c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
<h3 /><h3>The domain of T</h3>Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0
⇒ x² ≥ 25
⇒ x ≥ ±5
⇒ -5 ≤ x ≤ 5.
<h3>Inverse image of f(x)</h3>
The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
f : R → R defined by f(x) = 3x³
Let f(x) = y.
So, y = 3x³
Dividing through by 3, we have
y/3 = x³
Taking cube root of both sides, we have
x = ∛(y/3)
Replacing y with x we have
y = ∛(x/3)
Replacing y with f⁻¹(x), we have
So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)
<h3>Inverse image of g(x)</h3>
The inverse image of g(x) is
g : R+ → R defined by g(x) = ln(x).
Let g(x) = y
y = ln(x)
Taking exponents of both sides, we have
Replacing x with y, we have
Replacing y with g⁻¹(x), we have
So, the inverse image of g(x) is
The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
h : R → R defined by h(x) = x − 9
Let y = h(x)
y = x - 9
Adding 9 to both sides, we have
y + 9 = x
Replacing x with y, we have
x + 9 = y
Replacing y with h⁻¹(x), we have
So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
Learn more about inverse image of a function here: