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Lera25 [3.4K]
3 years ago
12

Find the sum of a finite geometric sequence from n = 1 to n = 6, using the expression −2(5)n − 1.

Mathematics
2 answers:
vazorg [7]3 years ago
8 0
\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=-2\\
r=5\\
n=6
\end{cases}
\\\\\\
S_6=-2\left( \cfrac{1-5^6}{1-5} \right)
ehidna [41]3 years ago
5 0

The answer is D. −7,812

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Alexxx [7]

Answer:

2/7 of the waffles did not have blueberries or chocolate chips.

Step-by-step explanation:

So first, it says 4/7 were blueberry, and 1/7 were chocolate chip. When you add them two you get 5/7. The denominator is 7, so subtract 7 from 5 and you get 2, 2/7. 2/7 of the waffles did not have blueberries or chocolate chips.

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3 years ago
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pentagon [3]

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Step-by-step explanation:

4 0
3 years ago
Simplify 7^-5/6*7^-7/6
DochEvi [55]

So the rule with multiplying exponents of the same base is x^m*x^n=x^{m+n} . Apply this rule here:

7^{-\frac{5}{6}}*7^{-\frac{7}{6}}=7^{-\frac{5}{6}+{-\frac{7}{6}}}=7^{-\frac{12}{6}}=7^{-2}

Next, the rule with converting negative exponents into positive ones is x^{-m}=\frac{1}{x^m} . Apply this rule here:

7^{-2}=\frac{1}{7^2}=\frac{1}{49}

<u>Your final answer is 1/49.</u>

<h2>------------------------------------------------</h2>

So an additional rule when it comes to exponents is x^{\frac{m}{n}}=\sqrt[n]{x^m}

In this case, your fractional exponent, x^9/7, would be converted to \sqrt[7]{x^9} . However, I had just realized you can further expand this.

Remember the rule I had mentioned earlier about multiplying exponents of the same base? Well, you can apply it here:

\sqrt[7]{x^9}=\sqrt[7]{x^7*x^2}=x\sqrt[7]{x^2}

Your final answer would be x\sqrt[7]{x^2}

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3 years ago
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Answer:34

Step-by-step explanation:

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3 years ago
What is the classification for this polynomial 2u^2vw^2
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It’s a monomial for sure
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4 years ago
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