Question

va rog ajutor plizzzzzzzz

Answer 1

1. a ) 69  b)150  c) 6
2 ) 69 lei 
3) MK= $\frac{50}{100} * WK = \frac{1}{2} *30 =15 m$
SK=MK=15 m
SWK =$\frac{SK}{WK} = \frac{15}{30} = \frac{1}{2}$ => M ( ∡SKM)=180°-90°-30°=60 °
4) SI=[tex] \frac{1}{2} *WK= \frac{30}{2} =15 m [/tex]
$KZ= \frac{1}{2} *WK= \frac{30}{2} =15 m$
Deci vor ajunge in acelasi timp.

1) x= $\frac{14*3}{17*2} = \frac{21}{17}$ (simplificam 2 cu 14 .
2) 40 %
3) - (-50 ) + (14-45) =50 + (-31)= 19
b) -196 : (-7)²+961: (-31)-8 * (-2²)=-196 : 49 + )-31) -(-32) =-4-31+32= -35+32= -3

Answer 2

Partea II

1. 3/2x=17/14
x=21/17

2.10+15=25 elevi
15/25=3/5

3. a. -(-57+7)+(-7+21-45)=19
b.-196:(-7)^2+961:(-31)-8*(-2^2)=-3
-196 : 49 + )-31) -(-32) =-4-31+32= -35+32= -3

PARTEA I

4. Fie A locul in care se afla chiosculA este mijlocul segmentului [WK]
(WA)≡(AK)
Fie distantele de la W la A si de la K la A
VITEZA 1 = VITEZA 2
T1=D1/V1
T2=D2/V2=D1/V1
T1=T2
AJUNG IN ACELASI TIMP

3. MK= 50/100 * WA= 1/2*30=15
SK=MK=15 m
SWK = => M ( ∡SKM)=180°-90°-30°=60 °