Question

face si pt mine careva testu 15 va rog, dau 25 pct. ​

Answer 1

1) (4/5+3/4):2/1=

=(4/5+3/4)*1/2

=(16/20+15/20)*1/2

=31/20*1/2

=31/40

2) 20% din 80 = (80*20)/100 = 160/100=1,6

80+1.6= 81,6 lei

3) $\frac{x}{2} =\frac{18}{y} , x\neq 0\\x^{2} = 36\\x=±6\\x=-6\\x=6, x\neq 0\\x=-6\\x=6\\x1=-6,x2=6$

4. ABCD romb

AC=6 cm

O intersectia dreptelor AC si BD => AO = 4 cm

                                                           BO = 3 cm

in triunghiul AOB rezulta conform teoremei lui Pitagora ca AB²=AO²=OB²

16+9=25 => AB = 5 cm => AB = BC = CD = AD = 5 cm => P rombului ABCD = 5*4=20 cm

5. (ABC) ll (A´B´C´) => m(A´B´,BC) = m(AB, BC)=60 grade

Answer 2

Răspuns:

1. (4/5+0,75):2 =( 4/5+75/100):2

in paranteza trebuie sa aduci la acelasi numitor.. amplifici prima fractie cu 20 ... dupa ce aduci la acelasi numitor va arata asa⬇️⬇️⬇️

(80/100+75/100): 2 =155/100:2

acum simplicici fractia cu 5

31/20:2= 31/20 × 1/2= 31/40

2. t-tricoul

1t=80 lei

t + 20% × t

80+ 20/100×80

simplifica ce putem si rezulta asta:

80+2/1×8

80+2×8

80+16

96

3. x/2=18/×

2x=18×2

2×=36

x=36/2

x=18

4.

Aromb=d1×d2/2

Aromb=6×8/2

Aromb=48/2

Aromb=24cm²

5.

ABC si A'B'C'- planuri || =>m( A'B', BC)=m(AB BC)= m(ABC) = 60°(pentru ca ptisma triunghiulara regulata are baza un triunghi echilateral