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Aleks04 [339]
3 years ago
6

Which expression can be simplified as

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
7 0
The answer is (n^-3)^6
The fourth option
Ymorist [56]3 years ago
5 0
Step by step:
The answer is (n^-3)^6
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Write 77.5% as a fraction in simplest form.
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31/40 is it in a fraction
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Danika signs up to work for 3 1/2 hours at the science fair. If each work shift is 3/4 hour, how many shifts will Danika work?
forsale [732]

namely, how many times does 3/4 go into 3½?  Let's firstly convert the mixed fraction to improper fraction.

\bf \stackrel{mixed}{3\frac{1}{2}}\implies \cfrac{3\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{7}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{2}\div \cfrac{3}{4}\implies \cfrac{7}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{\stackrel{2}{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{3}\implies \cfrac{14}{3}\implies 4\frac{2}{3}

5 0
3 years ago
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Jose Rivera purchased 7 textbooks for his classes at River Run Community College. The total cost of the texts was $208.55. What
shepuryov [24]
In the question, there are certain information's that are of immense importance in regards to finding the answer. The first information is that Jose Rivera had purchased 7 textbooks for his classes at River Run Community College. The price that Jose Rivera had to pay was $208.55. The average cost per textbook needs to be found.
The cost of 7 textbooks purchased by Jose Rivera = 208.55 dollars
Then
The cost of 1 textbook purchased by Jose Rivera = (208.55/7) dollars
                                                                               = 29.79 dollars
The average cost of a textbook purchased by Jose Rivera rounded to the nearest cent is 29.80 dollars.
6 0
3 years ago
What is the value of the 9th term in the following geometric sequence?
earnstyle [38]

Answer:  The correct option is (D) 196608.

Step-by-step explanation:  We are given to find the value of the 9th term in the following geometric sequence :

3,     12,    48,    192,    .     .     .

We know that

the n-th term of a geometric sequence with first term a and common ratio r is given by

a_n=ar^{n-1}.

For the given sequence, we have

first term, a = 3  and the common ratio, r is given by

r=\dfrac{12}{3}=\dfrac{48}{12}=\dfrac{192}{48}=~~.~~.~~.~~=4.

Therefore, the 9th term of the given sequence will be

a_9=ar^{9-1}=3\times 4^8=3\times65536=196608.

Thus, the required 9th term of the given sequence is 196608.

Option (D) is CORRECT.

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3 years ago
Gracie went for a morning walk and met her old friend Suzie and her daughter Jennie after a long time. Gracie asked Suzie how ol
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