Answer:
D'G' = 52.5 units
Step-by-step explanation:
Since the dilatation is centred at the origin then multiply the original coordinates by the scale factor 3.5
D' = (1 × 3.5, 7 × 3.5 ) = (3.5, 24.5 )
G' = (- 8 × 3.5, - 5 × 3.5 ) = (- 28, - 17.5 )
Calculate D'G' using the distance formula
d =
with (x₁, y₁ ) = D' (3.5, 24.5) and (x₂, y₂ ) = G' (- 28, - 17.5)
D'G' =
=
=
=
= 52.5 units
Answer:
Step-by-step explanation:
In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:
a = -32 ft/s/s
v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)
Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)
t = ?? sec.
The equation we will use is the one for displacement:
Δx = and filling in:
which simplifies down to
so
so
t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)
Now we use that time in the x-dimension. Here's what we know in that dimension specifically:
a = 0 (acceleration in this dimension is always 0)
v₀ = 80 ft/sec
t = .968 sec
Δx = ?? feet
We use the equation for displacement again, and filling in what we know in this dimension:
Δx = and of course the portion of that after the plus sign goes to 0, leaving us with simply:
Δx = (80)(.968)
Δx = 77.46 feet