Based on the amount that Steve Weatherspoon wants to withdraw every year beginning in June 30, 2024, and the interest rate, the balance on June 30th 2023 should be $45,203.
<h3>What should the balance be in 2023?</h3>
The fact that Steve Weatherspoon wants to be able to withdraw a particular amount every year, this makes this amount an annuity.
The value in 2023 would therefore be the present value of the annuity that will then accrue to the required amounts as the years go by.
The present value of an annuity is:
= Annuity amount per year x Present value interest factor of an annuity, 11%, 3 years between 2024 and 2027
Solving gives:
= 13,126.25 x 3.44371
= $45,203
In conclusion, the balance on the fund in 2023 should be $45,203 in order for Steve Weatherspoon to achieve his objectives.
Find out more on the present value of an annuity at brainly.com/question/25792915
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Answer:
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Step-by-step explanation:
Answer:
X = 0, π/2 in the interval [0, 2pi).
Step-by-step explanation:
Use the auxiliary angle method:
R sin(x + a) = Rsin x cos a + Rcos x sin a = 1
sin x + cos x = 1
Comparing coefficients:
R cos a = 1 and R sin a = 1, so
tan a = R sin a / R cos a = 1
So a = π/4 radians.
Also R^2(sin^2 a + cos^2 a) = 1^2 + 1^2 = 2
Therefore R = √2.
So √2 sin (x +π/4 = 1
sin x + π/4 = 1/√2
x + π/4 = π/4
x = 0 radians
Also
x = 0 + π/2 = π/2.
Answer:
a. L{t} = 1/s² b. L{1} = 1/s
Step-by-step explanation:
Here is the complete question
The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0
Solution
a. L{t}
L{t} = ∫₀⁰⁰
Integrating by parts ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = 
and v = 
and du/dt = dt/dt = 1
So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w
So, ∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
∫₀⁰⁰ = []₀⁰⁰ - ∫₀⁰⁰
= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + 
[
]₀⁰⁰
= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]
= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]
= -1/s[(0 - 0] - 1/s²[0 - 1]
= -1/s[(0] - 1/s²[- 1]
= 0 + 1/s²
= 1/s²
L{t} = 1/s²
b. L{1}
L{1} = ∫₀⁰⁰
= []₀⁰⁰
= -1/s[exp(-∞s) - exp(-0s)]
= -1/s[exp(-∞) - exp(-0)]
= -1/s[0 - 1]
= -1/s(-1)
= 1/s
L{1} = 1/s
