Answer:
Smax = 676 ft
the maximum altitude (height) the rocket will attain during its flight is 676 ft
Step-by-step explanation:
Given;
The height function S(t) of the rocket as;
S(t) = -16t2 + 208t
The maximum altitude Smax, will occur at dS/dt = 0
differentiating S(t);
dS/dt = -32t + 208 = 0
-32t +208 = 0
32t = 208
t = 208/32
t = 6.5 seconds.
The maximum altitude Smax is;
Substituting t = 6.5 s
Smax = -16(6.5)^2 + 208(6.5)
Smax = 676 ft
the maximum altitude (height) the rocket will attain during its flight is 676 ft
Answer:
x3 + 4
Step-by-step explanation:
Scince Bill's friends amount of miles is unkown its x, and times it by three since it says times, and then add 4 since it says more than! Hope this helps! There is a really good website called Khan Academy and it explains everything! Just search in google your topic and say khan academy at the end and you will get practices and videos! Good luck!
Answer:
I) |xz| ≈ 28.6 km
II) |yz| ≈ 34.8 km
Step-by-step explanation:
Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)
|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°
Using Trigonometric ratio - SOHCAHTOA
I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20
|xz| = 20 * Tan 55 = 20 * 1.428
|xz| = 28.56 km
|xz| ≈ <u>28.6 km</u>
<u />
II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|
|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°
|yz| = 20 ÷ 0.574 = 34.84 km
|yz| ≈ <u>34.8 km</u>
Answer:
![\sqrt[n]{x^a}=x^{a/n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ea%7D%3Dx%5E%7Ba%2Fn%7D)
Step-by-step explanation:
While the "law"
![\sqrt[n]{x^n}=x](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5En%7D%3Dx)
may seem more applicable, and may seem to be a special case of the law shown in the answer above, it is not true in general. For example, ...
