What you want to do with this equation is make a triangle connecting the two points you have and count the distance (rise/run) for this you would get 6/1 !
Given a fragpole which makes a right angle with the point of observation with the angle of elevation as 50°.
Let x be the height of the flagpole, then
Therefore, the <span>equation that could be used to find the height of the flagpole is
</span><span>x/5 = sin 50</span>
Answer:
a) 150-90 = h(15.25 -9.75)
b) 10.91
Step-by-step explanation:
a) Jennifer has 150 -90 = 60 less in savings than Eduardo, but she is earning 15.25 -9.75 = 5.50 more per hour. An equation she can use to figure her hours is ...
60 = 5.50h . . . . for h hours worked.
In 'raw' form, that might be ...
150 -90 = (15.25 -9.75)h
__
b) Solving the above equation, we get ...
60/5.50 = h ≈ 10.90909 ≈ 10.91 . . . hours
Answer:
-72
Step-by-step explanation:
The answer is -72 because 8 * -9 = -72, I can't explain it to you any other way ;-;
Answer:
a) distance covered by hare d1 = 8t
b) distance covered by tortoise d2 = 5t + 550
c) ∆d = 550 - 3t
Step-by-step explanation:
Given;
Speed of hare u = 8m/s
Speed of tortoise v = 5 m/s
Initial distance of tortoise d0 = 550 m
a) using the equation of motion;
distance covered = speed × time + initial distance
d = vt + d0
For hare;
d0 = 0
Substituting the values;
d1 = 8t + 0
d1 = 8t
b)using the equation of motion;
distance covered = speed × time + initial distance
d2 = vt + d0
For tortoise;
d0 = 550m
Substituting the values;
d2 = 5t + 550
d2 = 5t + 550 m
c) the number of meters the tortoise is ahead of the hare.
∆d = distance covered by tortoise - distance covered by hare
∆d = d2 - d1
Substituting the values;
∆d = (5t + 550) - 8t
∆d = 550 - 3t
