III gimnazjum

Matematyka z plusem 3

a)2x+17=26−x   ∣+x    3x+17=26     ∣−17   3x=9       ∣:3   x=3

Rozwiązanie 1

Proporcje rozwiązujemy mnożąc na krzyż.   a)2x​=65​   6x=5⋅2   6x=10     ∣:6   x=610​=35​     b)2x​=4x+1​   4x=2(x+1)   4x=2x+2   ∣−2x    2x=2    ∣:2

a)L=2πr   ∣:2π    r=2πL​     b)P=21​ah   ∣⋅2   2P=ah    ∣:a   h=a2P​     c)h=2a3<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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