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 +
5z^2 + 16
a coefficient is a number multiplied by a variable (a letter). so there is only ONE coefficient in this expression and it is 5.
Answer:
A. 2086 yd²
Step-by-step explanation:
SA = 2lw + 2wh + 2lh
SA = 2*17*20 + 2*20*19 + 2*17*19
SA = 680 + 760 + 646
SA = 2086
Hope this helps :)
Answer:
The approximated length of EF is 2.2 units ⇒ A
Step-by-step explanation:
<em>The tangent ratio in the right triangle is the ratio between the opposite side to the adjacent side of one of the acute angle in the triangle</em>
In the given figure
∵ The triangle DFE has a right angle F
∵ The opposite side to angle D is EF
∵ The adjacent side to angle D is DF
→ By using the tangent ratio above
∴ tan(∠D) = 
∵ DF = 6 units
∵ m∠D = 20°
→ Substitute then in the ratio above
∴ tan(20°) = 
→ Multiply both sides by 6
∴ 6 tan(20°) = EF
∴ 2.183821406 = EF
→ Approximate it to the nearest tenth
∴ 2.2 = EF
∴ The approximated length of EF is 2.2 units
