Question

ma poate ajuta cineva ? va rog !
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Answer 1

8. x/5 = 4/3. Folosind proportiile derivate, avem: (x+5)/5 = (4+3)/3=7/3

9. Produsul in diagonala: ab=35. Deci ab/35=1

10. p=25. Reducere de 16%: 16*25/100 = 16/4=4 lei. Pretul nou: 25-4=21lei

11. Patratele de 2 cifre: 16, 25, 36, 49, 64, 81, deci 6 numee

Numerele de 2 cifre: 10, 11, ..., 99

Numarul total de numere de 2 cifre: (99 - 10) + 1= 90

Probabilitatea: 6/90 * 100 = 60/9 = 20/3 = 6.(6)%

12. (-3, 2] contine urmatotii intregi: -2, -1, 0, 1, 2. Deci suma lor este 0

13. [-5, 2) Contine 0. Orice numar inmultit cu 0 da 0. Deci produsul intregilor este 0

Answer 2

$\it 8)\ \ \dfrac{x}{5}=\dfrac{4}{3}\stackrel{derivare}{\Longrightarrow}\ \dfrac{x+5}{5}=\dfrac{4+3}{3}\Rightarrow \dfrac{x+5}{5}=\dfrac{7}{3}\\ \\ \\ 9)\ \ \dfrac{a}{5}=\dfrac{7}{b}\Rightarrow ab=35 \Rightarrow \dfrac{ab}{35}=\dfrac{35}{35}=1\\ \\ \\ 10)\ \ 25-\dfrac{16}{100}\cdot25=25-4=21\ lei$

$\it 11)\ \ p=\dfrac{nr.\ cazuri\ favorabile}{nr.\ cazuri\ posibile}\\ \\ \\ Cazuri\ favorabile:\ 16,\ 25,\ 36,\ 49,\ 64,\ 81.\\ \\ Avem\ 6\ cazuri\ favorabile.\\ \\ Cazuri\ posibile:\ 10,\ 11,\ 12,\ ...,\ 99\\ \\ Avem\ 99-9=90\ cazuri\ posibile.$

$\it p=\dfrac{\ 6^{(6}}{90}=\dfrac{1}{15}$

$\it 12)\ \ (-3,\ 2]\cap\mathbb{Z}=\{-2,\ -1,\ 0,\ 1,\ 2\}\\ \\ -2+(-1)+0+1+2=0\\ \\ 13)\ \ 0\in[-5,\ 2) \Rightarrow\ produsul\ cerut\ este\ egal\ cu\ \ 0.$