He area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is
<span>(x - x^2)^2 * sqrt(3)/4. </span>
<span>Integrating from x = 0 to x = 1, we have </span>
<span>[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4 </span>
<span>= (1/30)*sqrt(3)/4 = sqrt(3)/120 = about 0.0144. </span>
<span>Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...
</span>
Answer:
SOH-CAH-TOA
~~~~~~~~~~
H=
O= 5
A= 9
~~~~~~~~~~~~~~~~
Sin =
Cos=
Tan =
Step-by-step explanation:
Answer:
121
Step-by-step explanation:
10(3)^2-7(3)+10=121
Brillianst please!!!
24 numbers are inbetween 20 and 45.
The answer:
the vertex formula is V( -b/2a, f((-b/2a) )
so if The graph of y=ax^2+bx+c is a parabola that opens up and has a vertex at (0,5), then -b/2a = 0, implies b = 0, and when b =0 f((-b/2a = 0/2a=0))=f(0)
in our case, f(x)=<span>ax^2+bx+c, and it implies f(0) = c </span>≠ 5
so there is no value of a, b and c which verifies the equation of vertex, the set solution is {∅} (the empty set)