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4 years ago

Pease help me ASAP....................

Mathematics

2 answers:

Rasek [7]4 years ago

5 0

A.
$-5 +5(x+4)\\ -5 +5x+20\\ 5x+15$

b.
$3m+2(5m-7)\\ 3m+10m-14\\ 13m-14$

c.
$4(6t+9)-10t\\ 24t+36-10t\\ 14t+36$

d.
$8k-6(3-2k)\\ 8k-18+12k\\ 20k-18$

AfilCa [17]4 years ago

3 0

First one, x+4

second one, 13m-7

third one, 14x+36

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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

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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

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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 = $\frac{1}{3}$

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