Answer:
the value of painting increases each year. To find the value of the painting for the next year, the art dealer multiplies the current value by 1.6. if the original value of the painting is 100, what is the value of the painting next yearStep-by-step explanation:
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)
For Independent Events, P(A) × P(B) = P(A∩B)
so we have, P(A∩B) = 0.4×0.1 = 0.04
P(A') = 1 - 0.4 = 0.6
This information can be represented on a Venn diagram as shown below
P(A'∪B) means the union of everything that is not A with everything that is B
P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64
Answer:
w=4
Step-by-step explanation:
2(48)+2(8w)+2(6w)=208
1) Start by Distributing the value outside of the parenthesis:
96+16w+12w=208
2) Combine alike terms:
96+28w=208
3) Subtract 96 from both sides:
28w=112
4)Divide both sides by 28 to isolate w:
w=4
Let me know if you do not understand :)
Answer:
no solution
Step-by-step explanation:
n -8 = 6+n
Subtract n from each side
n-n -8 = 6+n-n
-8 = 6
This is never true so there is no solution