Question

cine ma ajuta cu astea, pls.?

Answer 1

Prima poza:
$\sqrt{1+2^0+2^1+2^2+...+ 2^{n-1}} =2^{15}$

ridicam la patrat:

$1+2^0+2^1+2^2+...+ 2^{n-1} =2^{30}$

$1+1+2^1+2^2+...+ 2^{n-1} =2^{30}$ - o inmultim cu 2

$2*1+2*1+2*2^1+2*2^2+...+ 2*2^{n-1} =2*2^{30}$

$2+2+2^2+2^3+...+ 2^{n-1}+2^n =2^{31}$

$(1+1+2+2^2+2^3+...+ 2^{n-1})+2^n =2^{31}$

$2^{30}+2^n =2^{31}$

$2^n =2^{31}-2^{30}$

$2^n =2^{30}(2-1)$

$2^n =2^{30}$

$n=30$

A doua poza:
Ducem AM_|_CD si BN_|_CD=> AM si BN sunt inaltimi in trapez
In triunghiurile dreptunghice ADM si BNC avem:
AD²=AM²+DM² ⇒5²=25=AM²+DM²
BC²=BN²+NC² ⇒32=BN²+NC²

scadem cele 2  si avem:
32-25=7=BN²+NC²-AM²-DM²

Dar:
AM=BN
si DM+NC=DC-AB=13-6=7

⇒7=NC²-DM²=(NC-DM)(NC+DM)
7=(NC-DM)*7
NC-DM=1 ⇒NC=1+DM
Dar DM+NC=7 ⇒NC=7-DM
⇒1+DM=7-DM
2*DM=6
DM=3

in ΔADM , avdm (din nou) :
AD²=AM²+DM²
⇒5²=AM²+3²
AM²=25-9=16
AM=4

Arie trapez= (AB+CD)*AM/2=(6+13)*4/2=19*2=38