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:
<em>LCM</em> = 
Step-by-step explanation:
Making factors of 
Taking
common:

Using <em>factorization</em> method:

Now, Making factors of 
Taking
common:

Using <em>factorization</em> method:

The underlined parts show the Highest Common Factor(HCF).
i.e. <em>HCF</em> is
.
We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

Using equations <em>(1)</em> and <em>(2)</em>:

Hence, <em>LCM</em> = 
Answer:
None of these
Step-by-step explanation:
The lines aren't parallels so none of these property's can be applied
Answer:
y=1/x-1/5
Step-by-step explanation:
tell me if u have more questions
y=-x+5
is the line so to make it perpendicular
y=1/x-1/5 i gtg
Answer:
△ABC and △DEF are similar by SAS Similarity Theorem.
Option (C) is correct .
Step-by-step explanation:
Definition of SAS Similarity property
Two triangles are said to be similar by SAS Similarity property If two sides are proportional and one corresponding angle congruent .
In △ABC and △DEF


and


Thus

∠ABC = ∠DEF (As given in the figure )
Thus △ABC and △DEF are similar by SAS Similarity property .