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kherson [118]
3 years ago
6

Which expression is equivalent to the following complex fraction? StartFraction 2 Over x EndFraction minus StartFraction 4 Over

y EndFraction divided by StartFraction negative 5 Over y EndFraction + StartFraction 3 Over x EndFraction StartFraction 3 y + 5 x Over 2 (y minus 2 x) EndFraction StartFraction 2 (y minus 2 x) Over 3 y minus 5 x EndFraction StartFraction 2 (y minus 2 x) (3 y minus 5 x) Over x squared y squared EndFraction StartFraction x squared y squared Over 2 (y minus 2 x) (3 y minus 5 x) EndFraction
Mathematics
1 answer:
aliina [53]3 years ago
8 0

Answer:

its A

Step-by-step explanation:

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Step-by-step explanation:

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What is the range of f(x) = (3/4)^x – 4?
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8b + 5 + 4 = -23 <br> Need ASAP
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Ayudemen porfavor . Determine 4 ala potencia x + 4 ala potencia -x, si se sabe que 2ala potencia x + 2 ala potencia -x =3
Afina-wow [57]

Answer:

Es igual a 7.

Step-by-step explanation:

Ok, sabemos que:

2^x + 2^{-x} = 3

Queremos calcular:

4^x + 4^{-x}

Tambien sabemos que 4 = 2*2

Entonces:

4^x + 4^{-x} = (2*2)^x + (2*2)^{-x}

y sabemos que 2*2 = 2^2

Entonces:

4^x + 4^{-x} = (2*2)^x + (2*2)^{-x} = (2^2)^x + (2^2)^{-x}

y ahora podemos usar la relación:

(a^n)^m = (a^m)^n = a^{m*n}

entonces:

(2^2)^x + (2^2)^{-x} = (2^x)^2 + (2^{-x})^2

Ahora podemos completar cuadrados sumando y restando el termino:

2*(2^x)*(2^{-x})

Asi tendremos:

(2^x)^2 + (2^{-x})^2 =  (2^x)^2 + (2^{-x})^2 + 2*(2^x)*(2^{-x}) - 2*(2^x)*(2^{-x}) = (2^x + 2^{-x})^2 -  2*(2^x)*(2^{-x})

Entonces de momento tenemos que:

4^x + 4^{-x} = (2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x})

Sabemos que el termino que esta dentro del parentesis es igual a 3.

Y tambien podemos usar la propiedad:

a^n*a^m = a^{n + m}

en el termino de la derecha.

Asi tendremos:

(2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x}) = (3)^2 - 2*2^{x + (-x)}  = 9 - 2 = 7

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3(4x-3)(x+6) is the answer
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