Well, we know the cyclist left the western part going eastwards, at the same time the car left the eastern part going westwards
the distance between them is 476 miles, and they met 8.5hrs later
let's say after 8.5hrs, the cyclist has travelled "d" miles, whilst the car has travelled the slack, or 476-d, in the same 8.5hrs
we know the rate of the car is faster... so if the cyclist rate is say "r", then the car's rate is r+33.2
thus

solve for "r", to see how fast the cyclist was going
what about the car? well, the car's rate is r + 33.2
Answer:
Option a.
Step-by-step explanation:
In the given triangle angle A is a right angle so triangle ABC is a right angled triangle.
Opposite side of right angle is hypotenuse. So, CB is hypotenuse.
From figure it is clear that CA is shorter that segment BA.
All angles are congruent to itself. So angle C is congruent to itself.
We know that, if an altitude is drawn from the right angle vertex in a right angle triangle it divide the triangle in two right angle triangles, then given triangle is similar to both new triangles.
So, triangle ABC is similar to triangle DBA if segment AD is an altitude of triangle ABC.
Therefore, the correct option is a.
Answer:
x=149
Step-by-step explanation:
pythagoras theorem dictates that a squared +b squared= c squared
in this case, c=x
51*51+140*140=x*x
2601+19600=x*x
22201=x*x
x=149
Answer:
$2.25 per ft² (I think)
Step-by-step explanation:
Subtract 4.9 from 6.4 to get 1.5
Multiply 6.4 by 1.5 and multiply 4.9 by 1.5
Then you get 9.6 and 7.35
Subtract 7.35 from 9.6 and get 2.25