Answer:
58.5 ft²
Step-by-step explanation:
The figure is made up of a square, triangle and rectangle. Thus, the area of the figure is the sum of the areas of the square, triangle and rectangle.
Let's find the area of each!
<u>Square</u>
Area of square= side ×side
Area
= 3(3)
= 9 ft²
<u>Triangle</u>
Area of triangle= ½ ×base ×height
Base= 11 ft
Height= 3 ft
Area
= ½(11)(3)
= 16.5 ft²
<u>Rectangle</u>
Area of rectangle= length ×breadth
Area
= 3(11)
= 33 ft²
Area of figure
= 9 +16.5 +33
=
Answer:
Domain : {x | all real numbers} ; Range: {y | y > 0}
Step-by-step explanation:
The function can be written as :
Now, since x is exponent so it can take any real values. So, its domain of f(x) is all real numbers
But value of f(x) can not be less than 1 because for x = 0 the value of f(x) is 1 and also for any values of x, the value of f(x) can never be less than 1
So, Range of f(x) is all real numbers greater than 0
Hence, Domain and Range of f(x) is given by :
Domain : {x | all real numbers} ;
Range: {y | y > 0}
Answer:
volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( x + b )² dx
Step-by-step explanation:
Given the data in the question and as illustrated in the image below;
R is in the region first quadrant with vertices; 0(0,0), A(a,0) and B(0,b)
from the image;
the equation of AB will be;
y-b / b-0 = x-0 / 0-a
(y-b)(0-a) = (b-0)(x-0)
0 - ay -0 + ba = bx - 0 - 0 + 0
-ay + ba = bx
ay = -bx + ba
divide through by a
y = x + ba/a
y = x + b
so R is bounded by y = x + b and y =0, 0 ≤ x ≤ a
The volume of the solid revolving R about x axis is;
dv = Area × thickness
= π( Radius)² dx
= π ( x + b )² dx
V = π ₀∫^a ( x + b )² dx
Therefore, volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( x + b )² dx