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zmey [24]
3 years ago
14

Find the 25th term of an arithmetic sequence whose first term is 12 and whose common difference is ‒6.

Mathematics
1 answer:
blondinia [14]3 years ago
8 0
C -132 is the answer
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Read 2 more answers
Someone help please!!!!
LUCKY_DIMON [66]

Answer:

x = -2

TU = 4

UB = 2

Step-by-step explanation:

you can add x^2 with 4x+10 and equate it to 6:

x^2 + 4x + 10 = 6

x^2 + 4x + 4

then u can use the roots formula : x = (-b ± √ (b2 - 4ac) )/2a

so it'll be   x = {-4±[√16 - 4(4)]}/2

x= -2

then u can substitute it and find TU and UB

TU= (-2)^2 = 4

UB= 4(-2)+10 = 2

5 0
2 years ago
For the function given state the period f(t) =6sin(3t-pi/6)-1
lapo4ka [179]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}

\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}


now, with that template in mind, let's take a peek at yours

\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}
8 0
3 years ago
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