PLEASE HELP!!!! I WILL MARK BRAINLIEST!!!!

Mathematics

1 answer:

Elena L [17]3 years ago

7 0

Answer:

Step-by-step explanation:

You might be interested in

swame and luigi were printing calendars swame used 21/2 cartridge, while luigi used 1 3/4 ink cartridges. how many more ink cart

nirvana33 [79]

I’m not sure if this is correct but the answer would be 8.75 well that’s what I got hope this helps and if it dead pls give me a thanks

8 0

4 years ago

#82 will give brainliest to best answer!

Fofino [41]

Answer:

$\frac{6}{k-6}$

Step-by-step explanation:

First, we can factor all of the following equations to turn that weird, huge looking thing into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ ÷ $\frac{(k-6)^2}{k(k-6)}$ × $\frac{6(k-10)}{k(k + 6)}$ . We know that division is simply multiplication by the reciprocal, so that whole equation will turn into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ × $\frac{k(k-6)}{(k-6)^2}$ × $\frac{6(k-10)}{k(k+6)}$ . Now we can cancel out some values if they are both in the numerator and denominator, which will turn that still huge looking thing into $\frac{6}{k-6}$ which is our final answer, as it cannot be simplified further.

Hope this helped! :)

7 0

2 years ago

Read 2 more answers

Ok ik that I do math for ppl a lot but if someone could plz just help me with theses math problems (if u say that u will help I’

hodyreva [135]

I will help you with your problem

7 0

3 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

4 years ago

Ga college readiness mathematics

igomit [66]

What is the question?????

5 0

3 years ago