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1 year ago

una fracion reducicion a su minima expresion en igual a 1/8 . si la suma de sus terminos es 72. hallar la diferencia entre ellos

Mathematics

1 answer:

Mandarinka [93]1 year ago

7 0

La diferencia entre el denominador y el númerador de la fracción es:

64 - 8 = 56.

<h3>¿Como encontrar la fracción?</h3>

Tendremos una fracción de la forma:

$\frac{a}{b}$

Tal que:

$\frac{a}{b} = \frac{1}{8}$

$a + b = 72$

Eso es un sistema de ecuaciones

Primero, podemos reescribir la primer ecuación como:

$8a = b$

Ahora podemos reemplazar "b" en la segunda ecuación por "8a", así obtenemos:

$a + 8a = 72\\\\9a = 72\\\\a = 72/9 = 8$

Ahora podemos obtener el valor de b:

$b = 8a = 8*8 = 64$

La diferencia entre a y b es:

$b - a = 64 - 8 = 56$

Sí quieres aprender m'as sobre sistemas de ecuaciones:

brainly.com/question/13729904

#SPJ1

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2 years ago

Brandon is playing hide-and-seek with Katy and Tony. Katy is hiding 63 feet south of Brandon, and Tony is hiding due east of Kat

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Answer:

60 feets

Step-by-step explanation:

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2 years ago

A salsa recipe uses 14 ounces of peppers and 7 ounces of onions

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Answer:

$s = salsa$

$p = peppers$

$o = onions$

$s = 14p + 7o$

Step-by-step explanation:

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14 + 7 = 21 ounces.

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3 years ago

Simply the expression 7y+4x+4-2x+8

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3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

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Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago