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Anni [7]
3 years ago
12

You invest $9000 with a 6% interest rate compounded semiannually. After 9 yrs, how much money is in your account?

Mathematics
1 answer:
astraxan [27]3 years ago
7 0
\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$9000\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\

t=years\to &9
\end{cases}
\\\\\\
A=9000\left(1+\frac{0.06}{2}\right)^{2\cdot 9}
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A car is traveling up a steep ramp to a parking garage.the ramp is 445 feet long and rises a vertical distance of 80 feet. What
Alecsey [184]
With the information provided in the problem, we can create a right triangle with the ramp as its hypotenuse, and  the vertical rise as its opposite side from its angle of elevation.
Let E be the angle of elevation from the car to the end of the ramp. We now know that the hypotenuse of our triangle measures 445 feet, and the opposite side measures 80 feet, so we need a trig function that relates our angle of elevation with the hypotenuse and the opposite side. That function is sine:
sine(E)= \frac{80}{445}
E=arcsine( \frac{80}{445} )
E=10.36

We can conclude that the angle of elevation from the car to the end of the ramp is 10.36°.

3 0
3 years ago
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Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?
Naddik [55]

El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.

<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>

La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.

Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:

σ = W / (π · D² / 4)

Donde:

  • W - Peso de la caja, en newtons.
  • D - Diámetro del área transversal de la caja, en metros.

Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:

σ = 25 N / [π · (0.02 m)² / 4]

σ ≈ 79577.472 Pa

<h3>Observación</h3>

La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.

Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145

#SPJ1

5 0
1 year ago
I need help ,,, and tahnks☺
Mashcka [7]
Pythagora: 20^2=(4x)^2+(3x^2)
400=16x^2+9x^2
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x^2=16
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6 0
3 years ago
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What is the solution to the equation 1/4x +2= -5/8x - 5
nalin [4]

Answer:

x = -8

Step-by-step explanation:

1/4x +2= -5/8x - 5

Add 5/8x to each side

1/4x +5/8x +2= -5/8x+5/8x - 5

1/4x+5/8x +2=  - 5

Subtract 2 from each side

1/4x+5/8x +2-2=  - 5-2

1/4x+5/8x =  - 7

Get a common denominator

1/4 *2/2 x + 5/8x = -7

2/8x + 5/8x = -7

7/8x = -7

Multiply each side by 8/7

8/7x * 7/8x = -7 *8/7

x = -8

7 0
3 years ago
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denpristay [2]
I used the quadratic formula and got x= 4, -6
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3 years ago
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