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vichka [17]
3 years ago
13

When solving for a system of equations, if you get an answer of 6=0, what does this solution mean regarding the solution to the

system ?
Mathematics
2 answers:
lara [203]3 years ago
8 0
It means the system is inconsistent, which is another way to way, there's no solution, namely, both graphs do not intersect ever, and therefore there's no common point with such x,y pair for both.
mel-nik [20]3 years ago
6 0
It means that graph is inconsistent. Most likely it is a no solution
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Use a triple integral to find the volume of the tetrahedron T bounded by the planes x+2y+z=2, x=2y, x=0 and z=0
Tanzania [10]

Answer:

Volume of the Tetrahedron T =\frac{1}{3}

Step-by-step explanation:

As given, The tetrahedron T is bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0

We have,

z = 0 and x + 2y + z = 2

⇒ z = 2 - x - 2y

∴ The limits of z are :

0 ≤ z ≤ 2 - x - 2y

Now, in the xy- plane , the equations becomes

x + 2y = 2 , x = 2y , x = 0 ( As in xy- plane , z = 0)

Firstly , we find the intersection between the lines x = 2y and x + 2y = 2

∴ we get

2y + 2y = 2

⇒4y = 2

⇒y = \frac{2}{4} = \frac{1}{2} = 0.5

⇒x = 2(\frac{1}{2}) = 1

So, the intersection point is ( 1, 0.5)

As we have x = 0 and x = 1

∴ The limits of x are :

0 ≤ x ≤ 1

Also,

x = 2y

⇒y = \frac{x}{2}

and x + 2y = 2

⇒2y = 2 - x

⇒y = 1 - \frac{x}{2}

∴ The limits of y are :

\frac{x}{2} ≤ y ≤ 1 - \frac{x}{2}

So, we get

Volume = \int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}\int\limits^{2-x-2y}_{z=0} {dz} \, dy  \, dx

             = \int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{[z]}\limits^{2-x-2y}_0 {} \,   \, dy  \, dx

             = \int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{(2-x-2y)} \,   \, dy  \, dx

             = \int\limits^1_0 {[2y-xy-y^{2} ]}\limits^{1-\frac{x}{2}} _{\frac{x}{2} } {} \, \, dx

             = \int\limits^1_0 {[2(1-\frac{x}{2} - \frac{x}{2})  -x(1-\frac{x}{2} - \frac{x}{2}) -(1-\frac{x}{2}) ^{2}  + (\frac{x}{2} )^{2} ] {} \, \, dx

             = \int\limits^1_0 {(1 - 2x + x^{2} )} \, \, dx

             = {(x - x^{2}  + \frac{x^{3}}{3}  )}\limits^1_0

             = 1 - 1² + \frac{1^{3} }{3} - 0 + 0 - 0

             = 1 - 1 + \frac{1 }{3} =  \frac{1}{3}

So, we get

Volume =\frac{1}{3}

7 0
3 years ago
8*1,000,000+8*1/1,000,000
iren [92.7K]

Answer:

Step-by-step explanation:

8000000

<em>Hope that helps!</em>

8 0
3 years ago
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Elodia [21]

Answer: 2x

​2

​​ +5x−3

Step-by-step explanation:

2x

​2

​​ −x+6x−3

2x

​2

​​ +(−x+6x)−3

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3 years ago
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Find the product.<br><br> -7(-a^2)(-b^3)
Reil [10]
I belive the answer to your question would be -7a^2b^3

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4 years ago
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Simplify 8 to the 3rd power
stich3 [128]

Answer: 512

Step-by-step explanation: 8 x 8 x 8 = 512 (8x8=64, 64x8=512)

Hope this helps!

Mark Brainliest if you want!

8 0
3 years ago
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