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
A = b * h
b = 8 2/3
h = 5 3/4
A = 8 2/3 * 5 3/4.....turn to improper fractions
A = 26/3 * 23/4
A = 598/12
A = 49 5/6 ft^2 <====
Answer:
Step-by-step explanation:
x = 32
x - 10 = 22
add 10 to both sides of the equation
-10 + 10 = 0
22+ 10 = 32
x = 32
Answer:
1/4th Cup left
Step-by-step explanation:
1/2 cup is equal to 2/4 cup, so subtract that from 3/4th and you end with 1/4th
a² + 10a + 16
Find two numbers whose product equals 16 and sum equals 10:
<u>Product = 16</u> <u>Sum = 10</u>
1 * 16 1 + 16 = 17 <em>doesn't work</em>
2 * 8 2 + 8 = 10 THIS WORKS!
4 * 4 4 + 4 = 8 <em>doesn't work</em>
Since 2 & 8 satisfy our requirements, those are the values that go into the parentheses.
Answer: (a + 2)(a + 8)
Note: when there is a coefficient for a², a different method is used to find the factors.
