Almost got it!
x + 3 = 3(y + 2)/2 [Multiplied both sides by 3]
so x + 3 = (3y + 6)/2
2(x + 3) = 3y + 6 [Multiplied both sides by 2]
2x + 6 = 3y + 6
2x = 3y [Subtracted 6 from both sides]
x = 3y/2 [Divided both sides by 2]
x/3 = y/2 [Divided both sides by 3]
So you wrote y/3 instead of y/2
Hope this helped!
Before the driver applies the brakes ( with the reaction time ):
d 1 = v0 · t = 20 m/s · 0.53 s = 10.6 m
After that:
v = v0 - a · t1
0 = 20 m/s - 7 · t1
7 · t1 = 20
t1 = 2.86 s
d 2 = v 0 · t1 - a · t1² / 2
d 2 = 20 m/s · 2.86 s - 7 m/s² · (2.86 s)²/2 = 57.2 m - 28.6 m = 28.6 m
d = d 1 + d 2 = 10.6 m + 28.6 m = 39.2 m
Answer: the stopping distance of a car is 39.2 m.
Answer:
36 ÷ a
55 - b
18 + c
49 x d
e ÷ 62
Step-by-step explanation:
Answer
∴ The true statement is Answer:
The true statement is BD ≅ CE ⇒ 3rd answer
Step-by-step explanation:
- There is a line contained points B , C , D , E
- All points are equal distance from each other
- That means the distance of BC equal the distance of CD and equal
the distance of DE
∴ BC = CD = DE
- That means the line id divided into 3 equal parts, each part is one
third the line
∴ BC = 1/3 BE
∴ CD = 1/3 BE
∴ DE = 1/3 BE
∵ BC = CD
∴ C is the mid-point of BD
∴ BC = 1/2 BD
∵ CD = DE
∴ D is the mid-point of DE
∴ CD = 1/2 CE
- Lets check the answers
* BD = one half BC is not true because BC = one half BD
* BC = one half BE is not true because BC = one third BE
* BD ≅ CE is true because
BD = BC + CD
CE = CD + DE
BC ≅ DE and CD is common
then BD ≅ CE
* BC ≅ BD is not true because BC is one half BD
∴ The true statement is BD ≅ CE
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