Question

imi rezolva si mie cineva astea ?? bine .......ce stiti ! imi e lene sa le rezolv ! =)))))))))))))

689cd328f09ddfec74e62c8aad37ece2.pdf

c2bd691976e11ac71eae6157c8de9fbc.pdf

Answer 1

1.60
2.40 lei
3.{2}
4.25
5.36π
6.20

1.desenezi tu un paralelipiped
2.$media= \frac{2(4- \sqrt{7})+2 \sqrt{7} }{2}= \frac{8-2 \sqrt{7}+2 \sqrt{7} }{2}= \frac{8}{2} \\ media=4$
3.In prima zi 30/100 din x=3x/10
3x/10+350=x aduci la acelasi numitor
3x+3500=10x 
7x=3500 deci x=500km
4.f(x)=ax+3
f(-3)=0
-3a+3=0
-3a=-3 deci a=1
A(-3,0)
f(0)=b
0+3=b deci b=3, B(0,3)
Triunghiul ABC are OB=3u si AB=3u deci este isoscel.
5.
         (x+1)²-4        x²-x
E(x)= ------------ : -------- 
              x               x²

         (x+1-2)(x+1+2)         x²
E(x)= -------------------- · ---------
                   x                   x(x-1)

          x(x-1)(x+3)
E(x)= --------------- =x+3
              x(x-1)

E(m)=5
m+3=5 deci m=2

1.AD=30√3m
AB=30m
BM=10m deci AM=20m
a)A=AB*AD=30*30√3=900√3m²
b)In triunghiul ABC MN//AC deci BM/AB=BN/BC
10/30=BN/30√3 deci BN=10√3m⇒NC=20√3m=AQ
tg M=BN/MB=10√3/10=√3 deci M=60 grade
In triunghiul AMQ: tg M=AQ/AM=20√3/20=√3 deci M=60 grade
NMQ=180-60-60=60 grade
c)In triunghiul AMQ A=90 grade
AM²+AQ²=MQ²
20²+(20√3)²=MQ² deci MQ=√400+1200=40m
MQ//NP
MN//QP din amandoua ⇒MNPQ este paralelogram
deci MN=QP si MQ=NP
In triunghiul MBN B=90 grade
BM²+BN²=MN²
10²+(10√3)²=MN² deci MN=√100+300=20m
t=lungimea traseului
t=MN+NP+PQ+QM=20+40+20+40=120m

2.VO=h=12cm
OB=R
VA=15cm=G
a)R²+h²=G²
R²=15²-12²
R=√225-144=9cm=OB
V=πR²*h/3=81*12/3=324π cm³
b)sin (G,cerc)=sin(VB;OB)=sin VOB
sin VBO=VO/VB=12/15=3/4
c)As=9π deci r=3cm=O'B'
VO'/VO=VB'/VO-O'B'/OB
VO'/12=3/9=1/3
VO/=12/3=4cm