The net for a triangular prism consists of
2 identical triangles, here all nets have sides 5,12, and 13.
3 rectangles, all with a common dimension: the height H=14 of the prism.
Each of the 3 rectangles should have dimensions
H=14 plus one of the following:
each of 5, 12, 13 corresponding to one side of the base (triangle).
So the dimensions of the rectangles are
5x14, 12x14, 13x14
And dimensions of the triangles are
5,12 and 13.
The total surface area is therefore
(5+12+13)*14 + 2*(5*12/2)
=420+60
=480.
There is only one net that satisfies this condition.
X=45, X=62
X=73 thank me i am your savior I gave you the answer say thank you now you know
Answer:
I think it's d
Step-by-step explanation: Because they have equal sides and length
y = ln x , 1 <= x <= 3, about x axis and n = 10, dy/dx = 1/ x
S = (b a) ∫ 2π y √( 1 + (dy/dx) ^2) dx
so our f(x) is 2π y √( 1 + (dy/dx) ^2)
(b - a) / n = / 3 = (3-1) / 30 = 1/15
x0 = 1 , x1 = 1.2, x2 = 1.4, x3 = 1.6 ....... x(10) = 3
So we have , using Simpsons rule:-
S10 = (1/15) ( f(x0) + 4 f(x1) + 2 f)x2) +.... + f(x10) )
= (1/15) f(1) + f(3) + 4(f(1.2) + f(1.6) + f(2) + f(2.4) + f(2.8)) + 2(f(1.4) + f(1.8) + f(2.2) + f(2.6) )
( Note f(1) = 2 * π * ln 1 * √(1 + (1/1)^2) = 0 and f(3) = 2π ln3√(1+(1/3^2) = 7,276)
so we have S(10)
= 1/15 ( 0 + 7.2761738 + 4(1.4911851 +
Since there are
in a circle, and central
, you can see that the part "sliced out" by the central angle is
of the whole circle.
This means that the length of the intercepted arc
is
of the circumference of the circle. So,
