Question

{x,y,z}direct proportionale cu {0,1,0,2,0,3} 2·x+4·y+3·z=100

Answer 1

{x,y,z } d.p { 0,1;0,2;0,3}
2*x + 4*y + 3*z = 100
x/0,1= y/0,2=z/0,3 = k
x= 0,1 k
y= 0,2 k
x= 0,3 k
2*0,1k + 4*0,2k + 3*0,3k = 100
1.9 k = 100
k= 100:19
k= 52,6316
x= 52,6316*0,1 = 5,26316
y= 52,6316 * 0,2 = 10 ,526316
z= 52,6316*0,3 = 15,78948