Punkt P () należy do wykresu funkcji dla:
A. a=6
B. a=3
C. a=
D. a=
Prosze o obliczenia ;)
a do potegi ³/₂=3√3/8
√a³=3√3/8 /²
a³=27/64
a=∛(27/64)
a=¾
odp. c]
x = 3/2
y = 3√3 / 8
y = a^x
3√3 / 8 = a^(3/2)
a^(3/2) = 3^(1 +1/2) / 2^3
a^(3/2) = 3^3/2 / 2^3
[a^(1/2) ]^3 = (3^(1/2) / 2]^3
a^(1/2) = 3^(1/2) / 2
√a = √3 / 2 /²
a = 3/4 --------- odpowiedź
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a do potegi ³/₂=3√3/8
√a³=3√3/8 /²
a³=27/64
a=∛(27/64)
a=¾
odp. c]
x = 3/2
y = 3√3 / 8
y = a^x
3√3 / 8 = a^(3/2)
a^(3/2) = 3^(1 +1/2) / 2^3
a^(3/2) = 3^3/2 / 2^3
[a^(1/2) ]^3 = (3^(1/2) / 2]^3
a^(1/2) = 3^(1/2) / 2
√a = √3 / 2 /²
a = 3/4 --------- odpowiedź