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siapa yg bisa...........​

Berikut ini adalah pertanyaan dari laurenciachyankdh pada mata pelajaran Matematika untuk jenjang Sekolah Dasar

Siapa yg bisa..




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siapa yg bisa...........​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Number 7
There were \boxed{\bf2^{29}} bytes sent altogether.

Number 8

  • (a) The initial number of thousands of bacteria is \boxed{\bf 8}.
    (In other words, we can say that the initial number of bacteria is 8 thousands of bacteria.)
  • (b) The number of thousands of bacteria in each group after n hours is \boxed{\bf 16}.
    (In other words, we can say that number of thousands of bacteria in each group after n hours is 16 thousands of bacteria.)

Number 9

  • (a) The values of 3^2, 3^3, {\dots}, 3^8, and 3^9are9, 27, 81, 243, 729, 2187, 6561, and 19683, respectively.
  • (b) The pattern of 3^nwhenn=1,2,3,{\dots}is3, 9, 7, 1.
     

Solution/Explanation

Number 7

The total number of bytes of data that were sent altogether can obtained by multiplying the number of junk mails by the number of bytes of data each of them contains. Solving it gives us
\begin{aligned}2^{18}\times2^{11}=2^{18+11}=\bf2^{29}\ bytes\end{aligned}
\blacksquare

Number 8

(a) “Initial” means n = 0. So, the initial number of thousands of bacteria equals to 2^{0+3}=\bf2^3=8.

(b) To find the number of thousands of bacteria in each group, we just need to divide the number of thousands of bacteria by the number of groups, both in terms of n. Solving it gives us

\begin{aligned}\frac{2^{n+3}}{2^{n-1}}&=2^{n+3-(n-1)}\\&=2^{n-n+3+1}\\&=\bf2^4=16\end{aligned}
\blacksquare

Number 9

(a) We list those values one by one.

\begin{aligned}3^2&=3\times3&&\!\!\!={\bf9}.\\3^3&=9\times3&&\!\!\!={\bf27}.\\3^4&=27\times3&&\!\!\!={\bf81}.\\3^5&=81\times3&&\!\!\!={\bf243}.\\3^6&=243\times3&&\!\!\!={\bf729}.\\3^7&=729\times3&&\!\!\!={\bf2187}.\\3^8&=2187\times3&&\!\!\!={\bf6561}.\\3^9&=6561\times3&&\!\!\!={\bf19683}.\\\end{aligned}

(b) From previous solution, we can get that the pattern of 3^nwhenn=1,2,3,{\dots}is3, 9, 7, 1.

\blacksquare

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Last Update: Sat, 08 Oct 22
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