The table shows data for a car stopping on a wet road. Stopping distance formula Determine the value of f, rounded to the nearest hundredth.

We have that the value of f, rounded to the nearest hundredth.

f_1=0.02

f_2=0.009

f_3=0.009

From the question we are told that

d(v)= \frac{2.15v^2}{64.4f}

Generally the equation for the f is mathematically given as

f= \frac{2.15v^2}{64.4*d(v)}

Therefore

For v=20 and f=38

f= \frac{2.15(20)^2}{64.4*(38)(20)}

f=0.02

For v=30 and f=86

f=\frac{2.15(30)^2}{64.4*(86)(30)}

f=0.01

For v=40 and f=153

f=\frac{2.15(40)^2}{64.4*(153)(40)}

f=0.009

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0.35

the next one is d(v) = 2.15v^2 / 22.54

Step-by-step explanation:

i just did this assignment. also thanks to the answer above me :)

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

4 and 5)16.97 or 12√2

6)15.59 or 9√3

7)y = 15.59 or 9√3     x =27

8)17.80