You would subtract 5x from y so y=5x-10
Honestly photo math works for everything
Answer:
Height of the streetlight ≈ 8 ft(nearest foot)
Step-by-step explanation:
The doc file displays the triangle formed from the illustration. x is the height of the street light. The distance from the gentle man to the street light is 10 ft. He has a height of 5.6 ft and the shadow formed on the ground is 24 ft long. The height of the street light can be calculated below.
The length of the tip of the shadow to the base of the street light is 34 ft. Similar triangle have equal ratio of their corresponding sides .
ab = 5.6 ft
The ratio of the base sides = 24/34
The ratio of the heights = 5.6/x
The two ratio are equal Therefore,
24/34 = 5.6/x
24x = 5.6 × 34
24x = 190.4
divide both side by 24
x = 190.4/24
x = 7.93333333333
x ≈ 8 ft
Height of the streetlight ≈ 8 ft(nearest foot)
Sadly, after giving all the necessary data, you forgot to ask the question.
Here are some general considerations that jump out when we play with
that data:
<em>For the first object:</em>
The object's weight is (mass) x (gravity) = 2 x 9.8 = 19.6 newtons
The force needed to lift it at a steady speed is 19.6 newtons.
The potential energy it gains every time it rises 1 meter is 19.6 joules.
If it's rising at 2 meters per second, then it's gaining 39.2 joules of
potential energy per second.
The machine that's lifting it is providing 39.2 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(2)(4) = 4 joules.
<em>For the second object:</em>
The object's weight is (mass) x (gravity) = 4 x 9.8 = 39.2 newtons
The force needed to lift it at a steady speed is 39.2 newtons.
The potential energy it gains every time it rises 1 meter is 39.2 joules.
If it's rising at 3 meters per second, then it's gaining 117.6 joules of
potential energy per second.
The machine that's lifting it is providing 117.6 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(4)(9) = 18 joules.
If you go back and find out what the question is, there's a good chance that
you might find the answer here, or something that can lead you to it.