The answer to this question is: B;5i
v₀ = initial velocity of the freight train while it approach a road crossing = 16 km/h = 16 (5/18) m/s = 4.44 m/s
v = final velocity of the freight train after it crosses a road crossing = 65 km/h = 65 x 5/18 m/s = 18.06 m/s
t = time to do so = 10 min = 10 x 60 sec = 600 sec
acceleration is given as
a = (v - v₀ )/t
a = (18.06 - 4.44)/(600)
a = 0.023 m/s²
a = 294
Answer:
use the distance formula,
by using it, prove the adjacent sides of the quadrilateral DEFG
hence, DEFG would be a rhombus
Answer:
AC ≈ 5.03
Step-by-step explanation:
We can solve the problem above using the trigonometric ratio, they are;
SOH CAH TOA
sin Ф = opposite / hypotenuse
cosФ= adjacent/ hypotenuse
tan Ф = opposite / adjacent
From the diagram above, in reference to angle B;
opposite =AC and adjacent =BC
Since we have opposite and adjacent, the best formula to use is
tanФ = opposite / adjacent
tan B = AC / BC
tan 40 = AC/ 6
Multiply both-side of the equation by 6
6× tan 40 = AC/ 6 × 6
At the right-hand side of the equation, 6 will cancel-out 6 leaving us with just AC
6×tan 40 = AC
5.034598 = AC
AC ≈ 5.03 to the nearest hundredths
Its the last one because 10 times 1 is 10 and 0 plus 10 still gives you 10
