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Sagot :
To solve the problem, we need to fill in the missing values in the tables by performing the necessary calculations.
### Part 1: Complete the first table
We begin by filling in the missing 'valor' using the provided 'Base' and 'Exponente'. Specifically, when both 'Base' and 'Exponente' are present, 'valor' is calculated as [tex]\( \text{Base}^{\text{Exponente}} \)[/tex]:
1. For Base = 40 and Exponente = 3:
[tex]\[ \text{valor} = 40^3 = 64000 \][/tex]
2. For Base = 6 and Exponente = 4:
[tex]\[ \text{valor} = 6^4 = 1296 \][/tex]
3. For Base = 90 and Exponente = 3:
[tex]\[ \text{valor} = 90^3 = 729000 \][/tex]
4. For Base = 98 and Exponente = 3:
[tex]\[ \text{valor} = 98^3 = 941192 \][/tex]
Now, after the calculations, the completed table is:
\begin{verbatim}
|---- Base ----| 8 | 2 | 10 | 40 | 6 | 10 | 73 | 90 | 98 | 80 | 60 |
|-- Exponente -| 2 | 5 | | 3 | 4 | | 2 | 3 | 3 | | |
|---- valor ---| 64 | 32 |100 |64000|1296 | |4900|729000|941192| 6400|3600 |
\end{verbatim}
### Part 2: Complete the second table
Next, we fill in the missing 'suadrado' and 'cubo' values for each 'Jumero' from 2 to 12:
For 'suadrado', which is the square of each 'Jumero' ([tex]\( \text{suadrado} = \text{Jumero}^2 \)[/tex]):
\begin{itemize}
\item [tex]\( \text{Jumero} = 2 \rightarrow 2^2 = 4 \)[/tex]
\item [tex]\( \text{Jumero} = 3 \rightarrow 3^2 = 9 \)[/tex]
\item [tex]\( \text{Jumero} = 4 \rightarrow 4^2 = 16 \)[/tex]
\item [tex]\( \text{Jumero} = 5 \rightarrow 5^2 = 25 \)[/tex]
\item [tex]\( \text{Jumero} = 6 \rightarrow 6^2 = 36 \)[/tex]
\item [tex]\( \text{Jumero} = 7 \rightarrow 7^2 = 49 \)[/tex]
\item [tex]\( \text{Jumero} = 8 \rightarrow 8^2 = 64 \)[/tex]
\item [tex]\( \text{Jumero} = 9 \rightarrow 9^2 = 81 \)[/tex]
\item [tex]\( \text{Jumero} = 10 \rightarrow 10^2 = 100 \)[/tex]
\item [tex]\( \text{Jumero} = 11 \rightarrow 11^2 = 121 \)[/tex]
\item [tex]\( \text{Jumero} = 12 \rightarrow 12^2 = 144 \)[/tex]
\end{itemize}
For 'cubo', which is the cube of each 'Jumero' ([tex]\( \text{cubo} = \text{Jumero}^3 \)[/tex]):
\begin{itemize}
\item [tex]\( \text{Jumero} = 2 \rightarrow 2^3 = 8 \)[/tex]
\item [tex]\( \text{Jumero} = 3 \rightarrow 3^3 = 27 \)[/tex]
\item [tex]\( \text{Jumero} = 4 \rightarrow 4^3 = 64 \)[/tex]
\item [tex]\( \text{Jumero} = 5 \rightarrow 5^3 = 125 \)[/tex]
\item [tex]\( \text{Jumero} = 6 \rightarrow 6^3 = 216 \)[/tex]
\item [tex]\( \text{Jumero} = 7 \rightarrow 7^3 = 343 \)[/tex]
\item [tex]\( \text{Jumero} = 8 \rightarrow 8^3 = 512 \)[/tex]
\item [tex]\( \text{Jumero} = 9 \rightarrow 9^3 = 729 \)[/tex]
\item [tex]\( \text{Jumero} = 10 \rightarrow 10^3 = 1000 \)[/tex]
\item [tex]\( \text{Jumero} = 11 \rightarrow 11^3 = 1331 \)[/tex]
\item [tex]\( \text{Jumero} = 12 \rightarrow 12^3 = 1728 \)[/tex]
\end{itemize}
Now, after the calculations, the completed table is:
\begin{verbatim}
|----- Jumero ----| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---- suadrado ---| 4 | 9 | 16| 25 | 36 | 49 | 64 | 81 |100 |121 |144 |
|------ Cubo -----| 8 | 27| 64| 125| 216| 343| 512| 729|1000|1331 |1728 |
\end{verbatim}
By following these steps methodically, we have successfully completed both tables.
### Part 1: Complete the first table
We begin by filling in the missing 'valor' using the provided 'Base' and 'Exponente'. Specifically, when both 'Base' and 'Exponente' are present, 'valor' is calculated as [tex]\( \text{Base}^{\text{Exponente}} \)[/tex]:
1. For Base = 40 and Exponente = 3:
[tex]\[ \text{valor} = 40^3 = 64000 \][/tex]
2. For Base = 6 and Exponente = 4:
[tex]\[ \text{valor} = 6^4 = 1296 \][/tex]
3. For Base = 90 and Exponente = 3:
[tex]\[ \text{valor} = 90^3 = 729000 \][/tex]
4. For Base = 98 and Exponente = 3:
[tex]\[ \text{valor} = 98^3 = 941192 \][/tex]
Now, after the calculations, the completed table is:
\begin{verbatim}
|---- Base ----| 8 | 2 | 10 | 40 | 6 | 10 | 73 | 90 | 98 | 80 | 60 |
|-- Exponente -| 2 | 5 | | 3 | 4 | | 2 | 3 | 3 | | |
|---- valor ---| 64 | 32 |100 |64000|1296 | |4900|729000|941192| 6400|3600 |
\end{verbatim}
### Part 2: Complete the second table
Next, we fill in the missing 'suadrado' and 'cubo' values for each 'Jumero' from 2 to 12:
For 'suadrado', which is the square of each 'Jumero' ([tex]\( \text{suadrado} = \text{Jumero}^2 \)[/tex]):
\begin{itemize}
\item [tex]\( \text{Jumero} = 2 \rightarrow 2^2 = 4 \)[/tex]
\item [tex]\( \text{Jumero} = 3 \rightarrow 3^2 = 9 \)[/tex]
\item [tex]\( \text{Jumero} = 4 \rightarrow 4^2 = 16 \)[/tex]
\item [tex]\( \text{Jumero} = 5 \rightarrow 5^2 = 25 \)[/tex]
\item [tex]\( \text{Jumero} = 6 \rightarrow 6^2 = 36 \)[/tex]
\item [tex]\( \text{Jumero} = 7 \rightarrow 7^2 = 49 \)[/tex]
\item [tex]\( \text{Jumero} = 8 \rightarrow 8^2 = 64 \)[/tex]
\item [tex]\( \text{Jumero} = 9 \rightarrow 9^2 = 81 \)[/tex]
\item [tex]\( \text{Jumero} = 10 \rightarrow 10^2 = 100 \)[/tex]
\item [tex]\( \text{Jumero} = 11 \rightarrow 11^2 = 121 \)[/tex]
\item [tex]\( \text{Jumero} = 12 \rightarrow 12^2 = 144 \)[/tex]
\end{itemize}
For 'cubo', which is the cube of each 'Jumero' ([tex]\( \text{cubo} = \text{Jumero}^3 \)[/tex]):
\begin{itemize}
\item [tex]\( \text{Jumero} = 2 \rightarrow 2^3 = 8 \)[/tex]
\item [tex]\( \text{Jumero} = 3 \rightarrow 3^3 = 27 \)[/tex]
\item [tex]\( \text{Jumero} = 4 \rightarrow 4^3 = 64 \)[/tex]
\item [tex]\( \text{Jumero} = 5 \rightarrow 5^3 = 125 \)[/tex]
\item [tex]\( \text{Jumero} = 6 \rightarrow 6^3 = 216 \)[/tex]
\item [tex]\( \text{Jumero} = 7 \rightarrow 7^3 = 343 \)[/tex]
\item [tex]\( \text{Jumero} = 8 \rightarrow 8^3 = 512 \)[/tex]
\item [tex]\( \text{Jumero} = 9 \rightarrow 9^3 = 729 \)[/tex]
\item [tex]\( \text{Jumero} = 10 \rightarrow 10^3 = 1000 \)[/tex]
\item [tex]\( \text{Jumero} = 11 \rightarrow 11^3 = 1331 \)[/tex]
\item [tex]\( \text{Jumero} = 12 \rightarrow 12^3 = 1728 \)[/tex]
\end{itemize}
Now, after the calculations, the completed table is:
\begin{verbatim}
|----- Jumero ----| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---- suadrado ---| 4 | 9 | 16| 25 | 36 | 49 | 64 | 81 |100 |121 |144 |
|------ Cubo -----| 8 | 27| 64| 125| 216| 343| 512| 729|1000|1331 |1728 |
\end{verbatim}
By following these steps methodically, we have successfully completed both tables.
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