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Delvig [45]
3 years ago
10

Find the product:

rmula1" title=" \frac{1}{8} \times \sqrt{8} " alt=" \frac{1}{8} \times \sqrt{8} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
lozanna [386]3 years ago
5 0

Answer:

≈ 0.35

Step-by-step explanation:

First simplify \sqrt{8} and \frac{1}{8} :

\sqrt{8} ≈ 2.83

\frac{1}{8} = 0.125

Now multiply them as shown in the original equation:

0.125 x 2.83 = 0.35375 ≈ 0.35

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what is 1/8 times 20, what is 4/5 times 3/8, what is 1 1/8 times 1/9, what is 3/4 times 1/3 times 2/5, all must be simplified pl
bezimeni [28]
I hope this helps you


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4 years ago
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Ayuda por favor
sergeinik [125]

A partir de la definición de razón y la teoría de semejanza entre triángulos, la razón del área del triángulo AMN y el área del cuadrilátero BMNC es equivalente a 1/3.

<h3>¿Cómo determinar la medida de un lado de un triángulo desconocido?</h3>

En este problema tenemos un sistema formado por dos triángulos <em>similares</em>, la semejanza entre los dos triángulos se debe a la colinealidad entre los segmentos de línea AP' (triángulo <em>pequeño</em>) y AP'' (triángulo <em>grande</em>), así como de los lados AM y AB, así como los lados AN y AC, así como los <em>mismos</em> ángulos en la <em>misma</em> distribución. (Semejanza Lado - Ángulo - Lado)

En consecuencia, obtenemos las siguientes proporciones:

AP'/AP'' = MN/BC = 1/2     (1)

Finalmente, la proporción entre el triángulo AMN y el cuadrilátero BMNC es:

\frac{AMN}{ABC - AMN} = \frac{\frac{1}{2}\cdot a \cdot \left(\frac{1}{2}\cdot h \right)}{\frac{1}{2}\cdot (2\cdot a) \cdot  h - \frac{1}{2}\cdot a \cdot \left(\frac{1}{2}\cdot h \right)} = \frac{\frac{1}{4}\cdot a\cdot h }{a\cdot h - \frac{1}{4}\cdot a \cdot h }

\frac{AMN}{ABC - AMN} = \frac{\frac{1}{4} }{\frac{3}{4} } = \frac{1}{3}

A partir de la definición de razón y la teoría de semejanza entre triángulos, la razón del área del triángulo AMN y el área del cuadrilátero BMNC es equivalente a 1/3.

Para aprender sobre triángulos semejantes: brainly.com/question/21730013

#SPJ1

3 0
2 years ago
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