Gina was baking cookies.A dozen cookies requires 1/6 of a cup of chocolate chips.She wants to use 1/2 of a cup.How many dozen co
1 answer:
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Answer:
There are a ton of answers. For instance, 1m by 159m.
Step-by-step explanation:
The formula for the area is , where
and
are constants.
The formula for perimeter is
The question wants another rectangle that has the same perimeter, but a smaller area.
The perimeter of the two fields has to be:
Thankfully, the question never specified how much smaller the area has to be. So let's go to the logical extreme and make it 1 meter wide. To conserve the perimeter, we'd have to add the removed 64 meters to the original 95 meters of length, resulting in a rectangle with a width of 1m and a length of 159m. This results in an area of exactly 159 . The original rectangle had an area of 6175
. Mission accomplished!
But why stop there? Let's go even further down this rabbit hole!
Instead of 1, why don't we take it several steps further and use 0.0000001? At that point, the length would end up being 159.9999999m.
The area would be 0.00001599999999 .
Let's go further!!
Width: 0.00000000000000000001
Length: 159.99999999999999999999
Area: 0.00000000000000000001599999999999999999999 (approximate).
One more, for old time's sake.
Width: 0.00000000000000000000000000000000000000000000000001m
Length: 159.99999999999999999999999999999999999999999999999999m
Area: 0.00000000000000000000000000000000000000000000000015999999999999999999999999999999999999999999999999999
At that point, there's so little area in the plot of land that you could claim it has no area!
Answer:
a. The distance the train travelled in the first hour is approximately 28.3 miles
b. The location of the train at 5:00 p.m. is 53 miles east, and 46 miles west
c. The location of the train at any given time by the function, f(t) = (-8 + 24·t, -10 + 15·t)
d. The train does not collide with the cyclist when the bike goes over the train tracks
Step-by-step explanation:
a. The given information on the train's motion are;
The location south the train is spotted = 10 miles south and 8 miles west
The time the observer spotted the train = 2:00 pm
The location the train is spotted at 3:00 p.m. = 5 miles north and 16 miles east
Therefore, the difference between the two times the train was spotted, t = 3:00 p.m. - 2:00 p.m. = 1 hour
Making use of the coordinate plane for the two locations the train was spotted, we have;
The initial location of the train = (-10, -8)
The final location of the train = (5, 16)
Therefore the distance the train travelled in the first hour is given by the formula for finding the distance, 'd', between two points, (x₁, y₁) and (x₂, y₂) as follows;
Therefore;
The distance the train travelled in the first hour, d = 3·√89 ≈ 28.3 miles
b. The speed of the train, v = (Distance travelled by the train)/Time
∴ v ≈ 28.3 miles/(1 hour) = 28.3 miles per hour
The speed of the train in the first hour, v ≈ 28.3 mph
The direction of the train, θ, is given by the arctangent of the slope, 'm', of the path of the train;
∴ m = tan(θ) = (5 - (-10))/(16 - (-8)) = 0.625
c. Distance = Velocity × Time
At 5:00 p.m., we have;
The time difference, Δt = 5:00 p.m. - 3:00 p.m. = 2 hours
The distance, d₁ = (28.3 mph × 2 hours = 56.6 miles
Using trigonometry, we have the horizonal distance travelled, 'Δx', in the 2 hours is given as follows;
Δx = d₁ × cos(θ)
∴ Δx = 56.6 × cos(arctan(0.625)) ≈ 48
The increase in the horizontal position of the train, relative to the point (5, 16), Δx ≈ 48 miles
The vertical distance increase in the two hours, Δy is given as follows;
Δy = 56.6 × sin(arctan(0.625)) ≈ 30
The increase in the vertical position of the train, relative to the point (5, 16), Δy ≈ 30 miles miles
Therefore; the location of the train at 5:00 p.m. = ((5 + 48), (16 + 30)) = (53, 46)
The location of the train at 5:00 p.m. = 53 miles east, and 46 miles west
c. The function, 'f', that would give the train's position at time-t is given as follows;
The P = f(28.3·t, θ)
Where;
28.3·t = √(x² + y²)
θ = arctan(y/x)
Parametric equations
y - 5 = 0.625·(x - 16)
∴ y = 0.625·x - 10 + 5
The equation of the train's track is therefore, presented as follows;
y = 0.625·x - 5
d = 28.3·t
The y-component of the velocity, = 3*√89 mph × sin(arctan(0.625)) = 15 mph
Therefore, we have;
y = -10 + 15·t
The x-component of the velocity, vₓ = 3*√89 mph × cos(arctan(0.625)) = 24 mph
Therefore, we have;
x = -8 + 24·t
The location of the train at any given time, 't', f(t) = (-8 + 24·t, -10 + 15·t)
d. The speed of the cyclist next to the observer at 2:00 p.m., v = 10 mph
The distance of the cyclist from the track = The x-intercept = 5/0.625 = 8
The distance of the cyclist from the track = 8 miles
The time it would take the cyclist to react the track, t = 8 miles/10 mph = 0.8 hours
The location of the train in 0.8 hours, is f(0.8) = (-8 + 24×0.8, -10 + 15×0.8)
∴ f(0.8) = (11.2, 2)
At the time the cyclist is at the track along the east-west axis, at the point (8, 0), the train is at the point (11.2, 2) therefore, the train does not collide with the cyclist when the bike goes over the train tracks.