The answer is the option d, which is: d) 
The explanation for this problem is shown below:
1. Smplify the denominator and rewrite the numerator in this form:
2. Multiply the denominator and the numerator by the conjugated 
and simplify the expression, as following:
3. As you can see, you obtain the expression shown in the option mentioned above.
The sequence is decreasing by 3 each time
So the formula would be y =27-3(x-1)
Plug in 43
Y =27-3(43-1)
Y = 27-3(42)
Y = 27-126
Y = -99
The 43rd term of the sequence would be -99
Hope this helps :)
we can pretty much split the middle part into two trapezoids. Check the picture below.
so we really have one trapezoid and one square, each twice, so simply let's get the area of the trapezoid and sum it up with the area of the square, twice, and that's the area of the shape.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{\textit{parallel sides}}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=3\\ b=7 \end{cases}\implies A=\cfrac{5(3+7)}{2}\implies A=25 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{sum of areas}}{[25+(3\cdot 3)]}\cdot \stackrel{twice}{2}\implies [34]2\implies \underset{in^2}{68}](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%7B%5Ctextit%7Bparallel%20sides%7D%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D5%5C%5C%20a%3D3%5C%5C%20b%3D7%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B5%283%2B7%29%7D%7B2%7D%5Cimplies%20A%3D25%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20areas%7D%7D%7B%5B25%2B%283%5Ccdot%203%29%5D%7D%5Ccdot%20%5Cstackrel%7Btwice%7D%7B2%7D%5Cimplies%20%5B34%5D2%5Cimplies%20%5Cunderset%7Bin%5E2%7D%7B68%7D)
Answer:
Final answer is approx 6.644 years.
Step-by-step explanation:
Answer:
Step-by-step explanation:
No. Each side must be less than the sum of the remaining two sides.
10 ≮ 5+1
