Let slant height be s
Height = 38yd. = h
Side = 42yd. = a
S² = h² + (a/2)²
= 38² + 21²
= 1444 + 441
= 1885
S = √1885
S = 43.42yd.
Lateral area = 1/2 × 4a × s
= 2 × 42 × 43.42
= 3646.99
= 3647 yd.²
Answer:
63 is greater than 48.1 ...
63 > 48.1
Step-by-step explanation:
63 is greater than 48 remember. So you need to remove the [.1] and focus on the 63 and 48.
<h3>Hope it helps!!</h3><h3><em>Please</em><em> </em><em>mark me as the brainliest</em><em>!</em><em>!</em><em>!</em></h3>
<em>Thanks</em><em>!</em><em>!</em><em>!</em><em>!</em><em>❤</em><em>❣</em><em>❤</em>
<h2>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>:)</h2>
Acute usually means below or exactly 90°, so the greatest would be 90
Answer:
x = 18
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
x+x+x=54
(x+x+x)=54(Combine Like Terms)
3x=54
3x=54
Step 2: Divide both sides by 3.
3x 3
=
54
3
x=18
Answer:
x=18
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)
