It is important to notice firs of all that the train with no stop acceleration and them no stop acceleration would have a faster travel between stops. given this is safe to assume that there is a period during while the train is not accelerating nor decelerating. To find the time that is speed is constant
let's assume non-stop accelerating and decelerating and the time lost (subtraction of both) is the time the train had constant speed:
assuming same time accelerating a decelerating we have that the distance is
x = voT + 1/2aT^2 (initiial speed = 0) (0.25 miles = 1320 feets)
1320 = 1/2 (8)(T/2)^2
T/2 = 18.17 seg
T = 36.34 seg
time of constant speed
T = 41 -36.34 = 4.66 seg
maximum speed
v = v0 + aT (initial speed = 0)
v = 8(18.17) (time accelerating)
v = 145.36 feets/sec
distance at full speed
x = v*t
x = 145.36*4.66 = 677.38 feets
M= 1/2
to find this you plug in the points into the the formula m=y2-y1 / x2-x1
m=3-6 / -4-2
m=-3 / -6
this simplifies to m=1/2
Answer:
steps below
Step-by-step explanation:
f(x) = -2/5 * (x+1)² + 7
Parent function: f(x) = x²
Transformation: Vertical shift up 7 units and horizontal shift left 1 unit
reflect about the x axis (+) : parabola faced downward
compression:(+) *2/5
Answer:
The answer is given below
Step-by-step explanation:
From the diagram below,Let the line AB and CD be parallel line. Let transversal line EF cut AB at Y and transversal line EF cut CD at Z.
The bisector of ∠BYZ and ∠DZY meet at O. Therefore ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2
∠BYZ and ∠DZY are interior angles.
∠BYZ + ∠DZY = 180 (sum of consecutive interior angles)
∠BYZ/2 + ∠DZY/2 = 180/2
∠BYZ/2 + ∠DZY/2 = 90°
In ΔOYZ:
∠YZO + ∠ZYO + ∠YOZ = 180 (sum of angles on a straight line).
But ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2
∠DZY/2 + ∠BYZ/2 + ∠YOZ = 180
90 + ∠YOZ = 180
∠YOZ = 180 - 90
∠YOZ = 90°
Therefore Angle bisectors of the same side interior angles are perpendicular.
Answer:
58.74 cubic meters
Step-by-step explanation:
The function is actually 
Here x is the number of days and the corresponding value of "f(x)" would be the water REMAINING after x days.
Since, we want to figure out how much water (from 60 m^3) is remaining after 7 days, we simply have to plug in x = 7 into the function. Whatever number we get, is the AMOUNT OF WATER REMAINING AFTER 7 DAYS.
So,

Thus, the amount of water remaining in the pool after 7 days is 58.74 cubic meters