---
# System prepended metadata

title: Innova 2S
tags: [' 2020', Innova, ' 2do Secundaria']

---

---
tags: Innova, 2do Secundaria, 2020
---

{%hackmd uDPEliYESW6lSHPBuyJ46A %}

# Innova 2S

![](https://i.imgur.com/qFEgndd.png)

<span class="caja">Solución</span>

$$
\begin{align*}
&= \sqrt{\frac{5}{7} \times \frac{343}{125}} \\
&= \sqrt{\frac{5 \times 7^3}{7 \times 5^3} } \\
&= \sqrt{\frac{7^2}{5^2} } \\
&= \sqrt{\left ( \frac{7}{5} \right)^2} \\
&= \frac{7}{5}
\end{align*}
$$

![](https://i.imgur.com/uDpd66R.png)

<span class="caja">Solución</span>

$$
\begin{align*}
&= \sqrt{16 \cdot 3} - 2\sqrt{4 \cdot 3} \\
&= 4\sqrt{3} - 2\cdot 2 \sqrt{3} \\
&= 4\sqrt{3} - 4 \sqrt{3} \\
&= 0
\end{align*}
$$


![](https://i.imgur.com/jAviywC.png)

<span class="caja">Solución</span>

$$
\begin{align*}
&= \sqrt{9 \cdot 5} - 3\sqrt{25 \cdot 5} \\
&= 3\sqrt{5} - 3\cdot 5 \sqrt{5} \\
&= 3\sqrt{5} - 15 \sqrt{5} \\
&= -12\sqrt{5}
\end{align*}
$$

![](https://i.imgur.com/QPDdwuF.png)

<span class="caja">Solución</span>
$$
\begin{align*}
&= \dfrac{a ^{\dfrac{5}{3}}}{ a^{\dfrac{1}{2}} } \\[10pt]
&= a ^{\dfrac{5}{3} - \dfrac{1}{2}} \\[10pt]
&= a ^{\dfrac{10-3}{6}} \\[10pt]
&= a ^{\dfrac{7}{6}} \\[10pt]
\end{align*}
$$

![](https://i.imgur.com/gjtWjF7.png)

<span class="caja">Solución</span>
$$
\begin{align*}
&= a ^{\dfrac{1}{3}} \cdot a^{\dfrac{7}{2}} \\[10pt]
&= a ^{\dfrac{1}{3} + \dfrac{7}{2}} \\[10pt]
&= a ^{\dfrac{2+21}{6}} \\[10pt]
&= a ^{\dfrac{23}{6}} \\[10pt]
\end{align*}
$$

![](https://i.imgur.com/u4eOkbY.png)

<span class="caja">Solución</span>

$$
\begin{align*}
M &= \frac{ \sqrt[5]{(5^2)^3} \cdot \sqrt[15]{5} \cdot \sqrt[3]{5^2} }{\sqrt[6]{5^2} \cdot \sqrt[5]{5^3}} \\[10pt]
M &= \frac{ 5^{\frac{6}{5}} \cdot 5^{1 \over 15} \cdot 5^{2 \over 3} }{ 5^{2 \over 6} \cdot 5^{3 \over 5} } \\[10pt]
M &= \frac{ 5^{ \frac{6}{5} + \frac{1}{15} + \frac{2}{3} } }{5 ^{\frac{1}{3} + \frac{3}{5}} } \\[10pt]
M &= \frac{ 5^{\frac{18 + 1 + 10}{15}} }{5^{\frac{5 + 9}{15}}} \\[10pt]
M &= \frac{ 5^{\frac{29}{15}} }{5^{\frac{14}{15}}} \\[10pt]
M &= 5^{\frac{29}{15} - \frac{14}{15}} \\[10pt]
M &= 5^{\frac{15}{15}} = 5^1 = 5 \\[10pt]
\end{align*}
$$

---

![](https://i.imgur.com/HUWZG3B.png)

![](https://i.imgur.com/yxyGJdK.png)

![](https://i.imgur.com/8R70sDS.png)

<span class="caja">Solución</span>

Perímetro: $2[3n^2 + (5n+5)]$

Reduciendo:

$6n^2 + 10n + 10$


![](https://i.imgur.com/ziyvUqx.png)

<span class="caja">Solución</span>

**Primera pregunta:**

Borde interno: $2[(x-2) + (x-9)]$
$\qquad \qquad \qquad = 2[2x -11]$
$\qquad \qquad \qquad = 4x -22$

Borde externo: $2[(2x+5) + 2x]$
$\qquad \qquad \qquad = 2[4x + 5]$
$\qquad \qquad \qquad = 8x + 10$

Total del borde del jardín: $4x - 22 + 8x + 10$
$\qquad \qquad \qquad \qquad \qquad = 12x - 12$

**Segunda pregunta:**

Si $x=4$ entonces comprará: $12 \times 4 - 12$
$\qquad \qquad \qquad \qquad \qquad \quad = 48 - 12$
$\qquad \qquad \qquad \qquad \qquad \quad = 36$

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![](https://i.imgur.com/4TiYidw.png)

![](https://i.imgur.com/ZJgruhg.png)

![](https://i.imgur.com/UnJeceq.png)

![](https://i.imgur.com/7d9Q0Qa.png)

<span class="caja">Solución</span>

$= 20 \times 4 - 10$
$= 80 - 10$
$= 70$

![](https://i.imgur.com/gzewmMv.png)

<span class="caja">Solución</span>

$= \dfrac{(3)^2}{11 + 3}$

$= \dfrac{9}{14}$