Podane niżej liczby zapisz w postaci i wskaż taką liczbę naturalną m, dla której prawdziwe są nierówności
a 4
b 9
c 8
d 19
e 4
f 22
g 9
h 7
i 7
a) 3 p(2) = p(9)*p(2) = p(9*2) = p(18)
3 p(2) = około 4,24
m = 4 , bo 4 < 4,24 < 5 , więc 4 < p(18) < 5
b)
7 p(2) = p(49)*p(2) = p(98 )
m = 9 , bo 9 < p(98) < 10
c)
5 p(3) = p(25)*p(3) = p(75)
m = 8, bo 8 < p(75) < 9
d)
11 p(3) = p(121)* p(3) = p (363)
m = 19, bo 19 < p(363 ) < 20
e)
2 p(5) = p(4)*p(5) = p(20)
m = 4 , bo 4 < p(20) < 5
f)
10 p(5) = p(100)*p(5) = p(500)
m = 22, bo 22 < p(500 ) < 23
g)
3 p(11) = p(9)*p(11) = p(99)
m = 9, bo 9 < p(99) < 10
h)
2 p(15) =p(4) *p(15) = p(60)
m = 7, bo 7 < p(60) < 8
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a 4
b 9
c 8
d 19
e 4
f 22
g 9
h 7
i 7
a) 3 p(2) = p(9)*p(2) = p(9*2) = p(18)
3 p(2) = około 4,24
m = 4 , bo 4 < 4,24 < 5 , więc 4 < p(18) < 5
b)
7 p(2) = p(49)*p(2) = p(98 )
m = 9 , bo 9 < p(98) < 10
c)
5 p(3) = p(25)*p(3) = p(75)
m = 8, bo 8 < p(75) < 9
d)
11 p(3) = p(121)* p(3) = p (363)
m = 19, bo 19 < p(363 ) < 20
e)
2 p(5) = p(4)*p(5) = p(20)
m = 4 , bo 4 < p(20) < 5
f)
10 p(5) = p(100)*p(5) = p(500)
m = 22, bo 22 < p(500 ) < 23
g)
3 p(11) = p(9)*p(11) = p(99)
m = 9, bo 9 < p(99) < 10
h)
2 p(15) =p(4) *p(15) = p(60)
m = 7, bo 7 < p(60) < 8
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