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 +
If you mean what is n, then n = 4
Answer:
The number of the cups of uncooked rice is 3 cups
The number of the cups of water is
cups
Step-by-step explanation:
* Let us read the recipe of Nasi Gorang
- 2 cups of uncooked rice need
cups of water to
make
cups of cooked rice
- Joseph needs
cups of cooked rice
- We need to know how many cups of uncooked rice Joshep needs
for this recipe and how much water should be added
* We can solve this problem by using the ratio method
⇒ uncooked cups : water cups : cooked cups
⇒ 2 :
: 
⇒ x : y : 
- By using cross multiplication
∵ x (
) = 2 (
)
∵
= 
∵
= 
∴ x (
) = 2 (
)
∴
x = 13
- Divide both sides by 
∴ x = 3
∴ The number of the cups of uncooked rice is 3 cups
- By using cross multiplication
∵ y (
) = (
)(
)
∵
= 
∵
= 
∵
= 
∴
y = (
)(
)
∴
y =
- Divide both sides by 
∴ y = 
∴ The number of the cups of water is
cups
Answer:
Step-by-step explanation:
Using the exponential growth function for the U. S. population from 1970 through 2003:
A = 205.1e^0.011t
with the U.S. population being 205.1 million in 1970, when would the U. S. population reach 350 million?
A.
2028
B.
2048
C.
2018
D.
2038
We have the expression
A = 205.1e^0.011t
We are asked to find when would the U. S. population reach 350 million
A = 350
350 = 205.1e^0.011t
We divide both sides by 205.1
Divide both sides by 205.1
350/205.1 = 205.1e^0.011t/205.1
1.7064846416 = e^0.011t
We take the log of both sides
log 1.7064846416 = log e^0.011t
log 1.7064846416 = t log 0.011
t = log 1.7064846416/ log 0.011
t = 0.1185057826