The angle is and acute angle.
Acute: Anything under 90 Degrees
Right: 90 Degrees
Obtuse: Anything Above 90 Degrees
Check the picture below.
so the picture has a rectangle that is 8 units high and 12 units wide, and it has a couple of "empty" trapezoids, with a height of 5 and "bases" of 9 and 3.
now, if we just take the whole area of the rectangle and then subtract the area of those two trapezoids, what's leftover is the blue area.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=9\\ b=3 \end{cases}\implies \begin{array}{llll} A=\cfrac{5(9+3)}{2}\implies A=30 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large Areas}}{\stackrel{rectangle}{(12\cdot 8)}~~ -~~\stackrel{\textit{two trapezoids}}{2(30)}}\implies 96-60\implies 36](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%20h%3D5%5C%5C%20a%3D9%5C%5C%20b%3D3%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B5%289%2B3%29%7D%7B2%7D%5Cimplies%20A%3D30%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20Areas%7D%7D%7B%5Cstackrel%7Brectangle%7D%7B%2812%5Ccdot%208%29%7D~~%20-~~%5Cstackrel%7B%5Ctextit%7Btwo%20trapezoids%7D%7D%7B2%2830%29%7D%7D%5Cimplies%2096-60%5Cimplies%2036)
(2x^2-5x)(4x^2-12x+10) = <span>8x^4 - 44x^3 + 80x^2 - 50x
So, the answer is option C.</span>
300words/min
1min= 60 secs
300words/60secs= 5
5words/second is your answer
= cos(x-π/2)/sin(x-π/2)
= (cos(x)cos(π/2) + sin((x)sin(π/2)) / (sin(x)cos(π/2) - cos(x)sin(π/2)
= sin(x)/ (-cos(x)) .... cos(π/2) = 0; sin(π/2) = 1
= -tan(x)
idk if this is right but i tried