X=36. You just make the equation: X+9=45. Then you subtract 9 from 45 like this: -9 -9. That gives you 36
Answer:
a. 4600
b. 6200
c. 6193
Step-by-step explanation:
Let
the number of elements in A.
Remember, the number of elements in
satisfies
![n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)](https://tex.z-dn.net/?f=n%28A_1%20%5Ccup%20A_2%20%5Ccup%20A_3%29%3Dn%28A_1%29%2Bn%28A_2%29%2Bn%28A_3%29-n%28A_1%5Ccap%20A_2%29-n%28A_1%5Ccap%20A_3%29-n%28A_2%5Ccap%20A_3%29-n%28A_1%5Ccap%20A_2%20%5Ccap%20A_3%29)
Then,
a) If
, and if ![A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000](https://tex.z-dn.net/?f=A_2%5Csubseteq%20A_3%2C%20n%28A_2%5Ccap%20A_3%29%3Dn%28A_2%29%3D1000)
Since ![A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1](https://tex.z-dn.net/?f=A_1%5Csubseteq%20A_2%5C%3B%20and%20%5C%3B%20A_2%5Csubseteq%20A_3%2C%20%5C%3B%20then%20%5C%3B%20A_1%5Ccap%20A_2%20%5Ccap%20A_3%3D%20A_1)
So
![n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600](https://tex.z-dn.net/?f=n%28A_1%20%5Ccup%20A_2%20%5Ccup%20A_3%29%3D%5C%5C%3Dn%28A_1%29%2Bn%28A_2%29%2Bn%28A_3%29-n%28A_1%5Ccap%20A_2%29-n%28A_1%5Ccap%20A_3%29-n%28A_2%5Ccap%20A_3%29-n%28A_1%5Ccap%20A_2%20%5Ccap%20A_3%29%3D%5C%5C%3D200%2B1000%2B5000-200-200-1000-200%3D4600)
b) Since the sets are pairwise disjoint
![n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200](https://tex.z-dn.net/?f=n%28A_1%20%5Ccup%20A_2%20%5Ccup%20A_3%29%3D%5C%5Cn%28A_1%29%2Bn%28A_2%29%2Bn%28A_3%29-n%28A_1%5Ccap%20A_2%29-n%28A_1%5Ccap%20A_3%29-n%28A_2%5Ccap%20A_3%29-n%28A_1%5Ccap%20A_2%20%5Ccap%20A_3%29%3D%5C%5C200%2B1000%2B5000-0-0-0-0%3D6200)
c) Since there are two elements in common to each pair of sets and one element in all three sets, then
![n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193](https://tex.z-dn.net/?f=n%28A_1%20%5Ccup%20A_2%20%5Ccup%20A_3%29%3D%5C%5C%3Dn%28A_1%29%2Bn%28A_2%29%2Bn%28A_3%29-n%28A_1%5Ccap%20A_2%29-n%28A_1%5Ccap%20A_3%29-n%28A_2%5Ccap%20A_3%29-n%28A_1%5Ccap%20A_2%20%5Ccap%20A_3%29%3D%5C%5C%3D200%2B1000%2B5000-2-2-2-1%3D6193)
Answer:
total area = 100
Step-by-step explanation:
10 x 4 = 40
1/2 * 6 * 10 = 30 ( x 2 triangle) = 60
total area = 40 + 60
total area = 100
Let x represent the side length of the square end, and let d represent the dimension that is the sum of length and girth. Then the volume V is given by
V = x²(d -4x)
Volume will be maximized when the derivative of V is zero.
dV/dx = 0 = -12x² +2dx
0 = -2x(6x -d)
This has solutions
x = 0, x = d/6
a) The largest possible volume is
(d/6)²(d -4d/6) = 2(d/6)³
= 2(108 in/6)³ = 11,664 in³
b) The dimensions of the package with largest volume are
d/6 = 18 inches square by
d -4d/6 = d/3 = 36 inches long