Answer:
Part A: From 0 to 2 seconds, the height of the water balloon increases from 60 to 75 feet, therefore the water balloon's height is increasing during the interval [0,2]
Part B: From 2 to 4 seconds, the height of the water balloon stays the same at 75 feet, therefore the water balloon's height is the same during the interval [2,4] From 10 to 12 seconds, the height of the water balloon stays the same at 0 feet, therefore the water balloon's height is the same during the interval [10,12] From 12 to 14 seconds, the height of the water balloon stays the same at 0 feet, therefore the water balloon's height is the same during the interval [12,14]
Part C: The interval, [4,6] of the domain is when the water ballon's height decreases the fastest. The interval [4,6] decreases by 35 feet. The two other intervals that decrease are [6,8] and [8,10] which both have the same slope. They decrease by 20 feet. Therefore, this helps us conclude that the interval [4,6] decreases the fastest because 35 feet is a more significant decrease than 20 feet.
Part D: I predict that the height of the water balloon at 16 seconds is 0 feet. This is because at 10-14 seconds, the water balloon's height is 0 feet. In read-world situations, if the water balloon is on the ground which is 0 feet, it stays on the ground due to gravity.
Step-by-step explanation:
I hope this helps! I also do not know if it is all correct but I did research and everything so hopefully it is correct! Good luck!
The resultant velocity of the plane is the sum of the two velocity vectors which are perpendicular to each other. See the attached figure.
The magnitude of the resultant velocity is
.
The approximate value of the actual velocity of the plane is
. Correct choice is (D).
<span> its D -The difference has 4 terms and a degree of 7.</span>
Answer:
Step-by-step explanation:
Answer:
a) y-8 = (y₀-8) , b) 2y -5 = (2y₀-5)
Explanation:
To solve these equations the method of direct integration is the easiest.
a) the given equation is
dy / dt = and -8
dy / y-8 = dt
We change variables
y-8 = u
dy = du
We replace and integrate
∫ du / u = ∫ dt
Ln (y-8) = t
We evaluate at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Let's simplify the equation
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) the equation is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
We integrate
½ Ln (2y-5) = t
We evaluate at the limits
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) the equation is very similar to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
Answer:
(5x²+2y² + 2xy)(5x²+2y²-2xy)
Step-by-step explanation:
25x⁴+4x²y²+4y⁴= (5x²)²+2*5x²*2y²+(2y²)² - 16x²y²= (5x²+2y²)² - (4xy)²= (5x²+2y² + 2xy)(5x²+2y² - 2xy)