The braking distance (dm) of a car is directly proportional to the square of its speed (v m/s)
When the speed of the car is increased by
50%, what is the percentage change in its
braking distance?

Step-by-step explanation:

We can assign variables as:

d  = original breaking distance

b  = original speed of the vehicle

d′  = increased breaking distance

b′  = increased speed of the vehicle

We can look at it this way, originally it was like this:

d=kb2  

Now, if we increase the speed to b’ which 200% more than b, we will have

d′=k(b′)2  

d′=k(b+2b)2  

d′=k(3b)2  

d′=9kb2  

Obtaining the difference between d and d’, we have

d′−d=9kb2−kb2  

d′−d=8kb2  

in terms of d

d′−d=8(kb2)  

d′−d=8d  

Adding d on both sides of the equation,

d′=d+8d  

Which means the original breaking distance is increased by 800%. This is similar to the phrase “the speed of the vehicle is increased by 200%” which translates to “b + 2b”.

Hope this answer helps you :)

Have a great day

Mark brainliest

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5

answer: translate the circles so they share a common center point and dilate circle y by a scale factor of 3.