a rocket is fired straight upward from ground level with an initial velocity of 480Km)hr. After t second , its distance above the ground level is given by 480t-16t^2. For what time interval is the rocket more than 3200Km above ground level?​

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west

19

step-by-step explanation:

12r - 8 - 12

you have to combine like terms. below i have bolded the like terms:

12r - 8 - 12

together it combines to:

12r - 20

hope this !