Answer:
see explanation
Step-by-step explanation:
a
The tangent and the radius at the point of contact form a right angle
Using Pythagoras' identity on the right triangle formed.
Let x be the distance from the centre to P, then
x² = 4² + 10² = 16 + 100 = 116 ( take the square root of both sides )
x = 
≈ 10.77 cm (to 2 dec. places )
b
let the required angle be Θ, then
Using the sine or cosine ratio in the right triangle.
cosΘ = 
= 
Θ = 
( ) ≈ 21.8°
Answer:
It's probably a^6
Step-by-step explanation:
a^2 * a^4
a^2+4
a^6
(-1, -5)
x-axis is to flip the y - coordinates
y-axis is to flip the x - coordinates
<span>12.3
Volume function: v(x) = ((18-x)(x-1)^2)/(4pi)
Since the perimeter of the piece of sheet metal is 36, the height of the tube created will be 36/2 - x = 18-x.
The volume of the tube will be the area of the cross section multiplied by the height. The area of the cross section will be pi r^2 and r will be (x-1)/(2pi). So the volume of the tube is
v(x) = (18-x)pi((x-1)/(2pi))^2
v(x) = (18-x)pi((x-1)^2/(4pi^2))
v(x) = ((18-x)(x-1)^2)/(4pi)
The maximum volume will happen when the value of the first derivative is zero. So calculate the first derivative:
v'(x) = (x-1)(3x - 37) / (4pi)
Convert to quadratic equation.
(3x^2 - 40x + 37)/(4pi) = 0
3/(4pi)x^2 - (10/pi)x + 37/(4pi) = 0
Now calculate the roots using the quadratic formula with a = 3/(4pi), b = -10/pi, and c = 37/(4pi)
The roots occur at x = 1 and x = 12 1/3. There are the points where the slope of the volume equation is zero. The root of 1 happens just as the volume of the tube is 0. So the root of 12 1/3 is the value you want where the volume of the tube is maximized. So the answer to the nearest tenth is 12.3</span>
