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k/7=n
b)a-6=k
c)y=7-z
d)d-a=bc/:b
d-a/b=c
e)x-1=2k/:2
x-1/2=k
f)2k-5=7n/:7
2k-5/:7=n
nie wiem czy dobrze, ale wydaję mi się, że tak ;)
k/7=n
b)a=k+6 |-6
a-6=k
c)z=7-y |-7
-y=z-7 |*(-1)
y=-z+7
d)d=a+bc |:b
d/b=a+c |-a
d/b-a=c
e)x=2k+1 |-1
x-1=2k |:2
(x-1)/2=k
f)2k=7n+5 |-5
2k-5=7n |:7
(2k/5)/7=n
k = 7n ||:7
n = k/7
b)
a = k + 6 ||-6
k = a - 6
c)
z = 7 - y ||+y
z + y = 7 ||-z
y = 7 - z
d)
d = a +bc ||-a
d - a = bc ||:b
c = (d - a) : b
e)
x = 2k + 1 ||-1
x - 1 = 2k ||:2
k = (x - 1) : 2
f)
2k = 7n + 5 ||-5
2k - 5 = 7n ||:7
n = (2k - 5) : 7
;))