Answer:
Step-by-step explanation:
c
I can figure out this is a frefall motion.
Starting from rest => Vo = 0
Then, use the equation: d = [1/2]gt^2 => t = √(2d/g)
d = width of a black/clear stripe pair = 5cm = 0.05m
g ≈ 10 m/s^2 (the real value is about 9.81 m/s^2)
t =√(2*0.05m/10m/s^2) = 0.1 s
Answer: approximately 0.1 s
Answer:
P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.
Step-by-step explanation:
P and Q are the intersection points of
x-y+1 = 0 and the circle x^2 + y^2 = 25
sub y = x+1 into the circle
x^2 + (x+1)^2 = 25
x^2 + x^2 + 2x + 1 - 25 = 0
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3 or x = -4
y = 4 or y = -3
so P(3,4) and Q(-4,3) are our two points
Height of triangle.
h = |0 - 0 + 1|/√2 = 1/√2
PQ = √( (-7)^2 + 1^2) = √50 = 5√2
area POQ = (1/2)(1/√2)(5√2) = 5/2 square units
hope this helped
Answer:
-5.6
Step-by-step explanation:
56/-10 = -5.6
<u>Answer-</u>
<em>The correct answer is</em>
<em>∠BDC and ∠AED are right angles</em>
<u>Solution-</u>
In the ΔCEA and ΔCDB,
As this common to both of the triangle.
If ∠BDC and ∠AED are right angles, then 
Now as
∠BCD = ∠ACE and ∠BDC = ∠AED,
∠DBC and ∠EAC will be same. (as sum of 3 angles in a triangle is 180°)
Then, ΔCEA ≈ ΔCDB
Therefore, additional information can be used to prove ΔCEA ≈ ΔCDB is ∠BDC and ∠AED are right angles.
