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
I don’t even know :( sorry
We can see that two sides of that triangle area equal
so, their corresponding angle will also be equal
now, we know that
sum of all angles in any triangle is always 180
so, we can get equation as

now, we can solve for x



so, option-C..............Answer
I think $8. That’s the difference between the highest and lowest number.
Answer:
–9t − 19t + 9t − –18t − t = 14
t = -7
Step-by-step explanation:
Hope this helps!!