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juin [17]
4 years ago
10

B) Arturo compró un automóvil usado y pago $2.500.000. Si este automóvil se devalúa (baja su precio) en un 20% anual. ¿Cuánto se

devalúa el primer año el precio del automóvil?
Mathematics
1 answer:
Radda [10]4 years ago
4 0

Answer:

$500.000

Step-by-step explanation:

Lo que debemos hacer es calcular el porcentaje de devaluación del precio total del automóvil, es decir calcular el 20% de $2.500.000, y eso lo podemos hacer de la siguiente manera:

2500000*0.2 = 500000

Lo que quiere decir que el primer año el automóvil de devaluó en $500.000

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For which inequality would x = 5 be a solution?
astra-53 [7]

Answer:

x + 7 ≥ 12

Step-by-step explanation:

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3 years ago
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The lengths of the sides of a triangle are in the ratio
Anton [14]
HOW TO SOLVE:

1) use x as a variable
2) the starting equation is:
          4x + 3x + 5x = 18

3) add all the like terms on the left side of the equation:
          12x = 18

4) Divide by 12 on both sides:
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5) Find the lengths of the sides:

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4 years ago
Use the picture attached.
Annette [7]

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2 years ago
1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial i
Katena32 [7]
Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be  2 - 5i and:

f(x)=\big(x-4\big)\big(x-(-8)\big)\big(x-(2+5i)\big)\big(x-(2-5i)\big)=\\\\=(x-4)(x+8)(x-2-5i)(x-2+5i)=\\\\=(x^2-4x+8x-32)(x^2-2x+5ix-2x+4-10i-5ix+10i-25i^2)\\\\=\big(x^2+4x-32\big)\big(x^2-4x+4-25\cdot(-1)\big)=\\\\=(x^2+4x-32)(x^2-4x+29)=\\\\=x^4-4x^3+29x^2+4x^3-16x^2+116x-32x^2+128x-928=\\\\=\boxed{x^4-19x^2+244x-928}

Answer B.
8 0
3 years ago
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1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t
natta225 [31]

The area of the surface given by \vec r_1(u,v) is 1. In terms of a surface integral, we have

1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv

By multiplying each component in \vec r_1 by 5, we have

\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}

and the same goes for the derivative with respect to v. Then the area of the surface given by \vec r_2(u,v) is

\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}

8 0
3 years ago
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