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4 years ago

Please help I really need to pass this exam If line GF is a midsegment of triangle CDE, find CD.

Mathematics

2 answers:

olga55 [171]4 years ago

3 0

CF = FG = GC because it is equilateral triangle

So
3x + 0.8 = 2x + 3

3x–2x = 3 – 0.8

X = 2.2

Putting this value to DE

IT IS ALSO A EQUILATERAL TRIANGLE

SO
5x + 4 = 5× 2.2 +4 =

11+4 = 15

All side will be 15 of triangle CD = DC = CE = 15

Mark as brainliest

brilliants [131]4 years ago

3 0

Answer:

$CD=13.6$

Step-by-step explanation:

We have been given a triangle CDE in which GF is mid-segment. we are asked to find the value of CD.

We know that the length of mid-segment of a triangle is half the measure of third side.

Using triangle mid-segment theorem, we can set an equation as:

$2(2x+3)=5x+4$

$4x+6=5x+4$

Switch sides:

$5x+4=4x+6$

$5x-4x+4=4x-4x+6$

$x+4=6$

$x+4-4=6-4$

$x=2$

Since F is midpoint of CD, so we can set an equation as:

$CD=2(3x+0.8)$

$CD=2(3(2)+0.8)$

$CD=2(6+0.8)$

$CD=2(6.8)$

$CD=13.6$

Therefore, the length of CD is 13.6 units.

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3 years ago

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3 years ago

What is the period and midline?

OverLord2011 [107]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}\qquad$

$\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}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\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}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

6 0

3 years ago