Check the picture below.
so the area of the hexagon is really just the area of two isosceles trapezoids.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ a=2\\ b=4\\ h=2 \end{cases}\implies \begin{array}{llll} A=\cfrac{2(2+4)}{2}\implies A=6 \\\\\\ \stackrel{\textit{twice that much}}{2A = 12} \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D2%5C%5C%20b%3D4%5C%5C%20h%3D2%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B2%282%2B4%29%7D%7B2%7D%5Cimplies%20A%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwice%20that%20much%7D%7D%7B2A%20%3D%2012%7D%20%5Cend%7Barray%7D)
We will use double angle identities:
cos (5x ) = sin (10x )
cos (5x ) = 2 cos (5x ) sin ( 5x )
cos ( 5 x) - 2 cos ( 5 x ) sin ( 5x ) = 0
cos ( 5 x ) · [ 1 - 2 sin (5 x) ] = 0
cos ( 5 x ) = 0 or : 1 - 2 sin (5 x) = 0
5 x = π/2 +kπ, k∈Z sin (5 x) = 1/2
x1 = π/10 + kπ/5 5 x = π/6+2kπ , k∈ Z
5 x = 5π/6 +2kπ , k∈ Z
x 2 = π/30 +2kπ/5
x 3 = π/9 + 2kπ/5
Answer:
30/100×$1500 = $450
Step-by-step explanation:
Tommy needs total capital of $1500 and he wants to lend 30% of loan from kaunderman, which is calculated as above.
Answer:
See solution below
Step-by-step explanation:
According to the diagram shown
m<1 = m<5 = 5=65 degrees (corresponding angle)
m<5 = m<4 - 65 degrees (alternate interior angle)
m<9 = m<8 = 65degrees (corresponding angle)
m<5 = m<8 = 65dgrees (vertically opposite angles)
m<6+m<8 = 180
m<6 + 65 = 180
m<6 = 180 - 65
m<6 = 115degrees
m<2 = m<6 = 115degrees (corresponding angles)
m<6 = m<7 = 115degrees (vertically opposite angles)
m<3 = m<7 = 115degrees(corresponding angle)
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