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3 years ago

-2x + 2y + 3z = 0 -2x - y + z = -3 2x + 3y + 3z = 5

Mathematics

1 answer:

lyudmila [28]3 years ago

8 0

Steps:

isolate " x " for -2x + 2y + 3z = 0 ; x = - ( - 3z - 2y ) / 2

Subsititute : x = - ( - 3z - 2y ) / 2

- 2 * ( - ( - 3z - 2y ) / 2 ) - y + z = - 3

2 * ( - ( - 3z - 2y ) / 2 ) + 3y + 3z =5

isolate " z " for : - 2 ( - ( - 3z - 2y ) / 2 ) - y+ z = - 3 ; z = - ( - 3 + 3y / 2 )

subsititute z = - ( - 3 + 3y / 2 ) :

2* ( - ( -3* ( - ( - 3 +3y / 2 ) - 2y / 2 ) + 3y + 3 ( - ( - 3 + 3y / 2 ) = 5

isolate y for 2 ( -( - 3 ) - 3 + 3y / 2 ) - 2y / 2 ) + 3y + 3 ( - ( - 3 + 3y / 2 ) = 5

y = 1

for z = - ( - 3+ 3y ) / 2

subsititute y = 1

z = - ( - 3 + 3 * 1 ) / 2

z = 0

For x = - ( - 3z - 2y ) / 2

subsititute y = 1

subsititute z = 0

x = - ( - 3 *0 - 2 * 1 ) / 2

x = 1

x = 1 , y = 1 , z = 0

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Step-by-step explanation:

see the attached figure to better understand the problem

step 1

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step 2

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substitute

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step 3

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we know that

The area of the region inside the circumcircle of the triangle but outside the triangle is equal to the area pf the circle minus the area of triangle

$A=(\frac{27}{4} \pi-81\frac{\sqrt{3}}{4})\ units^2$

Simplify

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