Suppose that the side length is s. Imagine that one side is level. Draw a vertical line through the top vertex perpendicular to the base (level side). Label this vertical line "h." Then h and s are related as follows:

√3 h

sin 60 degrees = ------- = --------

2 s

s√3

and so h = the height of the triangle = -----------

The area of this triangle is (1/2)(base)(height), which here is:

(1/2)(s/2)( s√3 /2), or

(s^2)√3

A = ---------------

8 0

2 years ago

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Determine the direction that this parabola opens y=2x^2-4x-6<br><br><br> Down or up

dimaraw [331]

Answer: The parabola opens upwards.

Step-by-step explanation:

6 0

2 years ago

Find the average rate of change of the function over the given interval

sattari [20]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }$

8 0

3 years ago