Proszę o rozwiązanie
y = (2/3) x + 5
a) A = (-8; 4)
a1 = a = 2/3
czyli
y = (2/3) x + b1
Podstawiamy -8 za x oraz 4 za y :
4 = (2/3)*(-8) + b1
4 = - 16/3 + b1
b1 = 4 + 16/3 = 12/3 + 16/3 = 28/3
Odp.
y = (2/3) x + 28/3
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b)
B = ( 9; 2)
(2/3)*a2 = - 1 => a2 = - 3/2
y = (-3/2) x + b2
Podstawiamy 9 za x oraz 2 za y:
2 = (-3/2)*9 + b2
2 = -27/2 + b2
b2 = 2 + 27/2 = 4/2 + 27/2 = 31/2
y = (-3/2) x + 31/2 lub y = -1,5 x + 15,5
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y = (2/3) x + 5
a) A = (-8; 4)
a1 = a = 2/3
czyli
y = (2/3) x + b1
Podstawiamy -8 za x oraz 4 za y :
4 = (2/3)*(-8) + b1
4 = - 16/3 + b1
b1 = 4 + 16/3 = 12/3 + 16/3 = 28/3
Odp.
y = (2/3) x + 28/3
=====================
b)
y = (2/3) x + 5
B = ( 9; 2)
(2/3)*a2 = - 1 => a2 = - 3/2
y = (-3/2) x + b2
Podstawiamy 9 za x oraz 2 za y:
2 = (-3/2)*9 + b2
2 = -27/2 + b2
b2 = 2 + 27/2 = 4/2 + 27/2 = 31/2
Odp.
y = (-3/2) x + 31/2 lub y = -1,5 x + 15,5
===============================================