D = rt. The first car's rate is r, and the other one, going faster, is r + 12. The time they travel is 2 hours. Since one is going faster than the other, he has gotten farther. But, regardless of that, the distance between them is 232. So the distance the first one travels, r*t, which 2r, plus the distance the second one travels, 2(r+12) = 232. 2r + 2r + 24 = 232. Solve for r. 4r = 208 and r = 52. The slower one goes 52 and the faster one goes 64
The answer will end up being c
U=2x+3, x=u/2-3/2
u^2+8u+11=0
u1=(-8+sqrt(64-44))/2=-4+sqrt(5)
u2=-4-sqrt(5)
x1=u1/2-3/2=-7/2+sqrt(5)/2
x2=u2-3/2=-7/2-sqrt(5)/2
Answer:
ΔABX and ΔEDX
Step-by-step explanation:
Given: AX ≅ EX
BX ≅ DX
SAS implies a congruent relation with respect to Side-Angle-Side between the two required triangle.
AX ≅ EX (given)
BX ≅ DX (given)
<BXA ≅ <DXE (vertical opposite angles)
Therefore,
ΔABX ≅ ΔEDX (Side-Angle-Side, SAS, congruence property)
