5x-4y при x=0; y=13; x=1,2; y=3,25

Question

Алгебра: 5x-4y при x=0; y=13; x=1,2; y=3,25

harrasova2000ozvpju · Accepted Answer

5x-7y

а) При x = 3/5 , y= - 4/7
5*3/5-7*(-4/7)=
=3-(-4)=3+4=7

5*3/5=5/1*3/5=3/1=3
7*(-4/7)=7/1*(-4/7)=-4/1=-4

б) при x = - 0,8 , y = 0,6
5*(-0,8)-7*0,6=
=-4-4,2=-8,2

5
0,8
40
0
-4,0

7
*0,6
42
0
4,2

2.
а) 7/39 - (1/13 - 2/3) и 7/39 + ( 2/3 - 1/3)
10/13 > 20/39

1/13 - 2/3=3/39-26/39=- 23/39
7/39-(-23/39)=7/39+23/39=30/39=10/13

2/3 - 1/3=1/3
7/39+1/3=7/39+13/39=20/39

б) 3/5 + 1/8:5/4 и (3/5 + 1.8) : 5.4
7/10 > 4/9

1/8:5/4=1/8*4/5=1/2*1/5=1/10
3/5+1/10=6/10+1/10=7/10

3/5 + 1.8=3/5+1 4/5=3/5+9/5=12/5=2 2/5
2 2/5:5,4=2 2/5:5 2/5=12/5:27/5=12/5*5/27=4/1*1/9=4/9

MOGZ ответил