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34kurt
3 years ago
10

6(3n+1)-3(-1+2n)=n+2

Mathematics
1 answer:
kap26 [50]3 years ago
7 0
First, you want to distribute the 6 and the -3. This turns you equation into 18n+6+3-6n=n+2. Next, put all the n's on one side and numbers on the other. 18n-6n-n=2-6-3. Then, combine. 11n=-7 -> n=-7/11. Hope this helps!
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Step-by-step explanation:

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Two kinds of crated cargo, A and B, are to be shipped by truck. Each crate of cargo A is 50 cubic feet in volume and weighs 200
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Answer:

The correct answer is option b)

50a+10b<1000; 200a+360b < 7200; a> 0; b>0

Step-by-step explanation:

We are given that Two kind of crated cargo namely A and B to be shipped by truck.

<u>Cargo A:</u>

Volume of each crate of cargo A = 50 cubic ft

Weight of each crate of cargo A = 200 pounds

Let number of crates of cargo A to be shipped = a

Total volume of 'a' crates of cargo A = 50a cubic ft

Total weight of 'a' crates of cargo A = 200a pounds

<u>Cargo B:</u>

Volume of each crate of cargo B = 10 cubic ft

Weight of each crate of cargo B = 360 pounds

Let number of crates of cargo B to be shipped = b

Total volume of 'b' crates of cargo B = 10b cubic ft

Total weight of 'b' crates of cargo B = 360b

Total volume allowed in the truck is 1000 cubic ft

Total volume of 'a' crates of Cargo A and Total volume of 'b' crates of Cargo B = 50a+10b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)

So, the inequality becomes

50a+10b ....... (1)

Total weight allowed (load limit) in the truck is 7200 pounds

Total weight of 'a' crates of Cargo A and Total weight of 'b' crates of Cargo B = 200a+360b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)

So, the inequality becomes

200a+360b ....... (1)

And number of crates of cargo A and B are always a positive number.

So, a > 0 and b > 0.

So, the correct answer is option b.

<em>b. 50a+10b<1000; 200a+360b < 7200; a> 0; b>0</em>

3 0
3 years ago
Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t
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Answer: The volume of largest rectangular box is 4.5 units.

Step-by-step explanation:

Since we have given that

Volume = xyz

with subject to x+2y+3z=9

So, let z=\dfrac{9-x-2y}{3}

So, Volume becomes,

V=xyz\\\\V=xy(\dfrac{9-x-2y}{3})\\\\V=\dfrac{9xy-x^2y-2xy^2}{3}

Partially derivative wrt x and y we get that

9-2x-2y=0\implies 2x+2y=9\\\\and\\\\9-x-4y=0\implies x+4y=9

By solving these two equations, we get that

x=3,y=\dfrac{3}{2}

So, z=\dfrac{9-x-2y}{3}=\dfrac{9-3-3}{3}=\dfrac{3}{3}=1

So, Volume of largest rectangular box would be

xyz=3\times \dfrac{3}{2}\times 1=\dfrac{9}{2}=4.5

Hence, the volume of largest rectangular box is 4.5 units.

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