Question

Doar la cele incercuite cu verde... si rezolvare completa !
Va rog..

Answer 1

1b)
2(5x+3)-3(2x-3)-4=7(5x-2)+3(9-3x)+5x
10x+6-6x+9-4=35x-14+27-9x+5x
4x+11=31x+13
27x=-2
x=-2/27
1c)
3(4x-3)+5(6x-7)-5x=9(x+10)+11(x-12)+4
12x-9+30x-35-5x=9x+90+11x-132+4
37x-44=20x-119
17x=79
x=79/17

2b)
(12x-5)/4+(2-x)/3=3-2x
36x-15+8-4x=36-24x
32x-7=36-24x
58x=43
x=43/58

2e)
(2-3x)/3+(2+5x)/4=2x-3
8-12x+6+15x=24x-36
14+3x=24x-36
21x=60
x=60/21=20/7

6c)
l=2√10
d=4√5
P=9√10
6d)
l=6√5
d=6√10
A=180

7) d=10 
100=2l²
l²=50
l=5√2
P=5√2x4=20√2
A=5√2*5√2=50

8) a=3 dm
d=50cm=5dm
b=√(5²-3²)=√25-9=√16=4
P=(3+4)*2=14
A=3*4=12 

9)
10:2=5
20:2=10
l=√(5²+10²)=√(25+100)=√125=5√5
P=5√5*4=20√5

10)
Atrapez=(B+b)*h/2
h=√(10²-8²)=√36=6
A=(24+8)*6/2=96
diagonala=√(6²+16²)=√(36+256)=√292=2√73

11)
latura neparalela=√(3²+4²)=√(9+16)=√25=5

12)
sin 30=4/AB
1/2=4/AB
AB=8
AD=√(8²-4²)=√(64-16)=√48=4√3
P=2*4√3+2*8=8√3+16
A=4*4√3 =16√3

3)  AD inaltime
cos 30=1/2
BD/10√3=√3/2
BD=15  ⇒ BC=30
P=10√3*2+30=20√3+30

4)QN=√([6²+(6√3)²]=√(36+108)=√144=12
diagonalele in dreptunghi se intretaie in parti egale
rezulta MO=OQ=MQ=6 ⇒ triunghiul MOQ echilateral deco m(<MOQ)=60 grade

6)triunghiul este dreptunghic deoarece
BC²=AB²+AC²
(5√3)²=(5√2)²+5²
75=50+25
A=5*5√2/2=25√2/2=12,5√2
h=25√2*2/5√3=50√2/5√3=10√6/3

7)
CE inaltime
triunghiul CEB isoscel deoarece dreptunghic si <B=45 grade
rezulta CE=EB=10
AB=8+10=18
A=(18+8)*10/2=130
AC=√(10²+8²)=√(100+64)=√164=2√41
DB=√(10²+18²)=√(100+324)=√424=2√106