<h3>The worth after 4 years is $ 680.24</h3>
<em><u>Solution:</u></em>
<em><u>The formula for compound interest, including principal sum, is:</u></em>

Where,
A = the future value of the investment
P = the principal investment amount
r = the annual interest rate (decimal)
n = the number of times that interest is compounded per unit t
t = the time the money is invested
From given,
n = 1 ( since interest is compounded annually)
p = 500
t = 4

<em><u>Substituting the values we get,</u></em>

Thus the worth after 4 years is $ 680.24
Step-by-step explanation:
The Taylor series expansion is:
Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!
f(x) = 1/x, a = 4, and n = 3.
First, find the derivatives.
f⁽⁰⁾(4) = 1/4
f⁽¹⁾(4) = -1/(4)² = -1/16
f⁽²⁾(4) = 2/(4)³ = 1/32
f⁽³⁾(4) = -6/(4)⁴ = -3/128
Therefore:
T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!
T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³
f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0. So we can eliminate the top left option. That leaves the other three options, where f(x) is the blue line.
Now we have to determine which green line is T₃(x). The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).
The bottom right graph is the only correct option.
Answer:
a. v(t)= -6.78
+ 16.33 b. 16.33 m/s
Step-by-step explanation:
The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=
=
. We now multiply both sides of the equation by the integrating factor.
μv' + μkv = μg ⇒
v' + k
v = g
⇒ [v
]' = g
. Integrating, we have
∫ [v
]' = ∫g
v
= 
+ c
v(t)=
+ c
.
From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have
9.55 = 9.8 × 15/9 + c
= 16.33 + c
c = 9.55 -16.33 = -6.78.
So, v(t)= 16.33 - 6.78
. m/s = - 6.78
+ 16.33 m/s
b. Velocity of object at time t = 0.5
At t = 0.5, v = - 6.78
+ 16.33 m/s = 16.328 m/s ≅ 16.33 m/s
The volume generated by rotating the given region
about OC is
<h3>
Washer method</h3>
Because the given region (
) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.
solution
We first find the value of x and y









![v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy](https://tex.z-dn.net/?f=v%3D%20%5Cpi%20%5Cint%5Climits%5E2_o%3D%20%5B%5Cfrac%7By%5E%7B2%7D%20%7D%7B4%7D%20-%20%5Cfrac%7By%5E%7B8%7D%20%7D%7B2%5E%7B8%7D%20%7D%7D%20%20%5D%20dy)
![v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]](https://tex.z-dn.net/?f=v%3D%20%5Cpi%20%5B%5Cint%5Climits%5E2_o%20%7B%5Cfrac%7By%5E%7B2%7D%20%7D%7B4%7D%20%7D%20%5C%2C%20dy%20-%20%5Cint%5Climits%5E2_o%20%7B%5Cfrac%7By%7D%7B2%5E%7B8%7D%20%7D%20%5E%7B8%7D%20%7D%20%5C%2C%20dy%20%5D)
![v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi](https://tex.z-dn.net/?f=v%3D%5Cpi%20%5B%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7By%5E%7B3%7D%20%7D%7B3%7D%20%20%5Cint%5Climits%5E2_0%20-%20%5Cfrac%7B1%7D%7B2%5E%7B8%7D%20%7D%20%20%5Cfrac%7By%5E%7Bg%7D%20%7D%7Bg%7D%20%5Cint%5Climits%5E2_o%5C%5Cv%3D%20%5Cpi%20%5B%5Cfrac%7B1%7D%7B12%7D%20%282%5E%7B3%7D%20-0%29-%5Cfrac%7B1%7D%7B2%5E%7B8%7D%2A9%20%7D%20%282%5E%7Bg%7D%20-0%29%5D%5C%5Cv%3D%20%5Cpi%20%5B%5Cfrac%7B2%7D%7B3%7D%20-%5Cfrac%7B2%7D%7Bg%7D%20%5D%5C%5Cv%3D%20%5Cfrac%7B4%7D%7Bg%7D%20%5Cpi)
A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095
Answer:
X= 15
Step-by-step explanation:
multiply 5 on x/5 and 9
divide 3 and 45
x=15