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)
Answer:
you are already in the hundredths
Step-by-step explanation:
5.48
5= the ones place
4= the tenths place
8= the hundredths
Answer:
Sine of A should be 5/13
Step-by-step explanation:
Hope this Helped
<span>
6 Find an exact value. sin 75°
</span>sin(A+B)=sin(A)cos(B)+cos(A)sin<span>(B)
</span>sin(45)=cos(45)=(2^0.5)/2 sin(30)=0.5 cos(30)=(3^0.5)/2
sin(45+30)=sin(45)cos(30)+cos(45)sin(30)=(6^0.5+2^0.5)/4
the answer is the letter d) quantity square root of six plus square root of two divided by four.
<span>
7. Find an exact value. sine of negative eleven pi divided by twelve.
</span>sin(-11pi/12) = -sin(11pi/12) = -sin(pi - pi/12) = -sin(pi/12) = -sin( (pi/6) / 2)
= - sqrt( (1-cos(pi/6) ) / 2) = -sqrt( (1-√3/2) / 2 ) = -(√3-1) / 2√2=(√2-√6)/4
the answer is the letter c) quantity square root of two minus square root of six divided by four.
<span>
8. Write the expression as the sine, cosine, or tangent of an angle. sin 9x cos x - cos 9x sin x
</span>
sin(A−B)=sinAcosB−cosAsinB
sin(9x−x)= sin9xcosx−cos9xsinx= sin(8x)
the answer is the letter c) sin 8x
<span>
9. Write the expression as the sine, cosine, or tangent of an angle. cos 112° cos 45° + sin 112° sin 45°</span>
cos(A−B)=cosAcosB<span>+sinA</span>sinB
cos(112−45)=cos112cos45<span>+sin112</span>sin45=cos(67)
the answer is the letter d) cos 67°
10. Rewrite with only sin x and cos x.
sin 2x - cos 2x
sin2x =
2sinxcosx<span>
cos2x = (cosx)^2 - (sinx)^2 = 2(cosx)^2 -1 = 1- 2(sinx)^2</span>
sin2x- cos2x=2sinxcosx-(1- 2(sinx)^2=2sinxcosx-1+2(sinx)^2
sin2x- cos2x=2sinxcosx-1+2(sinx)^2
<span>
the answer is the letter <span>
b) 2 sin x cos2x - 1
+ 2 sin2x</span></span>
Answer:
To figure out the common denominator for these fractions, I'll first need to factor that quadratic in the denominator on the right-hand side of the rational equation. This will also allow me to find the disallowed values for this equation. Factoring gives me:
x2 – 6x + 8 = (x – 4)(x – 2)
The factors of the quadratic on the right-hand side "just so happen" to be duplicates of the other denominators. This often happens in these exercises. (So often, in fact, that if you get completely different factors, you should probably go back and check your work.)
Step-by-step explanation:
