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Pavlova-9 [17]
3 years ago
14

Honestly I dont know how to do this just please help me

Mathematics
1 answer:
V125BC [204]3 years ago
7 0
F(x) = 3x - 2
f(8) = 3(8) - 2
f(8) = 24 - 2
f(8) = 22
f(-5) = 3(-5) - 2
f(-5) = -15 - 2
f(-5) = -17
f(8) - f(-5) = 22 - (-17)
f(8) - f(-5) = 39
Hope this helped! Good luck! :)
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What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?
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\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------

\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5
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