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Katena32 [7]
4 years ago
11

¿Qué hora es? Si el cuadrado de la mitad del

Mathematics
1 answer:
AVprozaik [17]4 years ago
4 0

Answer:

Si el cuadrado de la mitad del número de horas que faltan transcurrir del día coinciden con el número de horas transcurridas del día, son las 16:00 hs.

Step-by-step explanation:

Si el cuadrado de la mitad del número de horas que faltan transcurrir del día coinciden con el número de horas transcurridas del día, son las 16:00 hs.

Esto es así porque, como primera medida, la mitad de horas que faltan transcurrir del día no puede ser mayor a 4, puesto que 5 al cuadrado da como resultado 25, es decir, excede el número de horas que tiene un día.

Entonces, siguiendo con dicho razonamiento en sentido decreciente, tenemos que 4 al cuadrado da como resultado 16 (4 x 4). En este caso, 4 sería la mitad de horas que faltan transcurrir en el día, y 16 las horas ya transcurridas. Entonces, como 16 mas 8 da 24, y esa es la cantidad de horas que tiene el día, ésta es la opción correcta.

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The mass of a an atom of helium is 6.65 x 10^−24. Use scientific notation to express the mass of 250 helium atoms.
melisa1 [442]
<span>1.6625 x 10−21</span>

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Igoryamba

Answer:

(4x - 3)(5x - 8) = 20x^2 -47x + 24

Step-by-step explanation:

Given

(4x - 3)(5x - 8)

Required

Identify and correct error

The error is when the inner and outer terms are multiplied. i.e.

4x * -8 = -32x not 32x

-3 * 5x = -15x not 15x

So, the expression is:

(4x - 3)(5x - 8) = 20x^2 - 32x -15x + 24

(4x - 3)(5x - 8) = 20x^2 -47x + 24

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3 years ago
Solve to find x <br> log (3x+5) to base 5= 2log (1-3x) to base 25
Eduardwww [97]

Answer:

x = -2/ 3

Step-by-step explanation:

in order to cancel out the logs they should have common bases

\mathsf{ log_{5}(3x + 5) = 2 log_{25}(1 - 3x)  }

we can write 25 as 5²

\mathsf{ \implies  log_{5}(3x + 5) =  2 \: log_{ {5}^{2} }(1 - 3x)  }

we know that the reciprocal of the exponents of the bases are multiplied to the log

\mathsf{ \implies  log_{5}(3x + 5) =   \frac{1}{\cancel{2} } \times \cancel{2} \: log_{ 5}(1 - 3x)  }

and now since the logs have common bases

\mathsf{ \implies  \cancel { log_{5}}(3x + 5) = \cancel{ log_{ 5}}(1 - 3x)  }

we're left with

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<u>x = -2/ 3</u>

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3 years ago
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Answer:

B.0.25

Step-by-step explanation:

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