The solution would be like
this for this specific problem:
<span>V = ∫ dV </span><span>
<span>= ∫0→2 ∫
0→π/2 ∫ 0→ 2·r·sin(φ) [ r ] dzdφdr </span>
<span>= ∫0→2 ∫
0→π/2 [ r·2·r·sin(φ) - r·0 ] dφdr </span>
<span>= ∫0→2 ∫
0→π/2 [ 2·r²·sin(φ) ] dφdr </span>
<span>= ∫0→2 [
-2·r²·cos(π/2) + 2·r²·cos(0) ] dr </span>
<span>= ∫0→2 [
2·r² ] dr </span>
<span>=
(2/3)·2³ - (2/3)·0³ </span>
<span>= 16/3 </span></span>
So the volume of the
given solid is 16/3. I am hoping that these answers have satisfied
your query and it will be able to help you in your endeavors, and if you would
like, feel free to ask another question.
Answer:
act as if the ratio is a fraction 4:3 would be 4/3..
Step-by-step explanation:
Answer:
C.
Step-by-step explanation:
PEMDAS
-6 x 12p = -72p
-6 x -8n = 48n
48n - 47n = n
Check the picture below.
something worth noticing 
so, we're really graphing x+2, with a hole at x = 3, however, when x = 3, we know that f(x) = 5, but but but, when x = 3, x+2 = 5, so we end up with a continuous line all the way, x ∈ ℝ, because the "hole" from the first subfunction, gets closed off by the second subfunction in the piece-wise.
Use the quadractic formula
