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RUDIKE [14]
3 years ago
12

Valerie deposited $2000 into an IRA that grew to $5000 by retirement. If Valerie is currently in the 15% tax bracket and retired

, how much tax will she have to pay when she withdraws the entire $5000?
Mathematics
2 answers:
egoroff_w [7]3 years ago
8 0

Answer:

750

Step-by-step explanation:

sdas [7]3 years ago
7 0
The answer will be $750
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19, 12, 3, 6, 18, 20, 4, 4, 8, 12, 16, 20, 10, 1, 7, 8 1. Find the range of the data ​
german

Answer:

The highest observation =20

The lowest observation = 1

Range of the data = 20-1

= 19(ans)

8 0
3 years ago
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If f(x) = -x + 4 and g(x) = x2 - 5, then f(g(-2)) = ?
schepotkina [342]

Answer:

f(g(- 2)) = 5

Step-by-step explanation:

Evaluate g(- 2) then use this result to evaluate f(x)

g(- 2) = (- 2)² - 5 = 4 - 5 = - 1, then

f(- 1) = - (- 1) + 4 = 1 + 4 = 5

6 0
3 years ago
4^3-(84÷14)^2<br><br> Please help! Show work too! thank you!! &lt;3
ANEK [815]

Answer:

28

Step-by-step explanation:

4^{3}-(84\div14)^{2}

Calculate 4 to the power of 3 = 64

64-(\frac{84}{14} )^{2}

Divide 84 by 14 = 6

64-6^{2}

Calculate 6 to the power 2 = 36

64-36

=28

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4 0
3 years ago
-4x -2(8x =1)=-(2x - 10)
lubasha [3.4K]
-4x-2(8x=1)=-(2x-10)

-First, use the distributive property.
So distribute -2 to 8x=1  then distribute the negative - to (2x-10)
So it'd be:  -4x-16x=-2=-2x+10
Then combine like terms:
-20x=-2=-2x+10
-18x=-2=10
-18x=8
Divide -18x by 8
x=-2.25

7 0
3 years ago
The The Laplace Transform of a function , which is defined for all , is denoted by and is defined by the improper integral , as
guapka [62]

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} = ∫₀⁰⁰e^{-st}t

Integrating by parts  ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = e^{-st} and v = \frac{e^{-st}}{-s} and du/dt = dt/dt = 1

So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w

So,  ∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + \frac{1}{s} [\frac{e^{-st} }{-s}]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} = ∫₀⁰⁰e^{-st}1

= [\frac{e^{-st} }{-s}]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

6 0
3 years ago
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