So hmmm if you notice the tickmarks, the left-side is split in halves, and the right-side is also split in halves, that makes EF the midsegment of the trapezoid
thus
![\bf \textit{midsegment of a trapezoid}\\\\ m=\cfrac{base1+base2}{2}\qquad \begin{cases} base1, base2=\textit{parallel sides}\\ ----------\\ base1=x+8\\ base2=58\\ m=6x \end{cases} \\\\\\ 6x=\cfrac{(x+8)+(58)}{2}\implies 6x=\cfrac{x+8+58}{2}\implies 12x=x+66](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bmidsegment%20of%20a%20trapezoid%7D%5C%5C%5C%5C%0Am%3D%5Ccfrac%7Bbase1%2Bbase2%7D%7B2%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Abase1%2C%20base2%3D%5Ctextit%7Bparallel%20sides%7D%5C%5C%0A----------%5C%5C%0Abase1%3Dx%2B8%5C%5C%0Abase2%3D58%5C%5C%0Am%3D6x%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A6x%3D%5Ccfrac%7B%28x%2B8%29%2B%2858%29%7D%7B2%7D%5Cimplies%206x%3D%5Ccfrac%7Bx%2B8%2B58%7D%7B2%7D%5Cimplies%2012x%3Dx%2B66)
and surely you know how much that is
Subtract <span>x</span> from both sides
<span>5=2x−x
</span> Simplify <span><span>2</span><span>−x</span></span> to <span>x</span>
<span>5=x
</span>Switch sides
<span><span>x=5.</span></span>
Answer:
3 in
Step-by-step explanation:
The volume of the large sphere is ...
V = (4/3)πr³ = 288π . . . cubic inches
The volume of the smaller sphere is ...
288π -252π = 36π . . . cubic inches
The ratio of volumes of the smaller sphere to the larger is the cube of the ratio of radii. The radius of the smaller sphere will be ...
(6 in)∛(36π/(288π)) = (6 in)∛(1/8) = 3 in . . . . small sphere radius
Change the fractions so they have common denominators. For example. If adding 4/8 + 2/3, a common denominator of the two could be 24. Change both fractions to have a denominator of 24. To do this:
(4/8)×(3/3)= 12/24
(2/3)×(8/8)=16/24
Now that both fractions have denominators of 24, you can simply add the numerators to get your answer. in this case, it would be 28/24
Answer: 120 ways
Step-by-step explanation:
Let's label the routes a, b, c, d, and e. The goal is to come up with as many unique permutations as possible.
1 route: a=1. one factorial (1!) =1*1=1
2 routes: ab, ba=2. 2!=1*2=2
3 routes: abc, acb, bac, bca, cab, cba=6. 3!=1*2*3=6
4 routes: 4!=1*2*3*4=24
5 routes: 5!=1*2*3*4*5=120 ways