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3 years ago

In the figure, AB¯¯¯¯¯∥CD¯¯¯¯¯ and m∠2=60°. What is m∠6?

Mathematics

1 answer:

PolarNik [594]3 years ago

8 0

<span>mâ 6 = 60Â° Since both AB and CD are parallel to each other, equivalent angles made by the intersection of the third line are equal to each other. â 2 is the upper right hand angle of the intersection between AB and the third line and â 6 is also the upper right hand angle of the intersection between CD and third line.</span>

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If the second term of a geometric sequence of real numbers is -2 and the fifth term is 16 then what is the fourteenth term?

Agata [3.3K]

<h3>Given</h3>

A geometric sequence such that ...

$a_2=-2\quad\text{and}\quad a_5=16$

$a_{14}$

<h3>Solution</h3>

We can use the ratio of the given terms to find the common ratio of the sequence, then use that to find the desired term from one of the given terms. We don't actually need the common ratio (-2). All we need is its cube (-8).

$a_2=a_1r^{(2-1)}=a_1r^1\\a_5=a_1r^{(5-1)}=a_1r^4\\a_{14}=a_1r^{(14-1)}=a_1r^13=a_5r^9\\\\\dfrac{a_5}{a_2}=\dfrac{a_1r^4}{a_1r^1}=r^3=\dfrac{16}{-2}=-8\\\\r^9=\left(r^3\right)^3=(-8)^3=-512\\a_{14}=a_5(-512)\\\\a_{14}=-8192$

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3 years ago

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viva [34]

Answer:

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Step-by-step explanation:

To find the reciprocal, flip the fraction. The sign remains the same.

The reciprocal of 8/5 is 5/8.

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Step-by-step explanation:

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Sedaia [141]

$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
&& y={{ k }}x
\end{array}\\ \quad \\
$

and also

$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{ k}}}{x}
\end{array}
$

now, we know that V varies directly to T and inversely to P simultaneously
thus $\bf V=T\cdot \cfrac{k}{P}$

so $\bf V=T\cdot \cfrac{k}{P}\qquad 
\begin{cases}
V=42\\
T=84\\
P=8
\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k
\\\\\\
V=\cfrac{4T}{P}\qquad now\quad 
\begin{cases}
V=74\\
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