Answer:
- r(0) = <0, 100> . . . . . . . .meters
- r'(0) = <7.071, 7.071> . . . . meters per second
Step-by-step explanation:
<u>Initial Position</u>
The problem statement tells us we're measuring position from the ground at the base of the building where the projectile was launched. The initial horizontal position is presumed to be zero. The initial vertical position is said to be 100 meters from the ground, so (in meters) ...
r(0) = <0, 100>
<u>Initial Velocity</u>
The velocity vector resolves into components in the horizontal direction and the vertical direction. For angle α from the horizontal, the horizontal component of velocity is v₁·cos(α), and the vertical component is v₁·sin(α). For v₁ = 10 m/s and α = π/4, the initial velocity vector (in m/s) is ...
r'(0) = <10·cos(π/4), 10·sin(π/4)>
r'(0) ≈ <7.071, 7.071>
Answer:
2. There is also an irrelevant joke that 1 plus 1 equals window.
Step-by-step explanation:
Answer:
Step-by-step explanation:
wheres the expression
The main rule we use in this exercise is
and the more general case:
![( \sqrt[n]{a} ) ^{m} =( a^{ \frac{1}{n}} )^{m} = a^{ \frac{1}{n}*m}=a^{ \frac{m}{n}}](https://tex.z-dn.net/?f=%28%20%20%5Csqrt%5Bn%5D%7Ba%7D%20%20%29%20%5E%7Bm%7D%20%3D%28%20a%5E%7B%20%5Cfrac%7B1%7D%7Bn%7D%7D%20%29%5E%7Bm%7D%20%3D%20a%5E%7B%20%5Cfrac%7B1%7D%7Bn%7D%2Am%7D%3Da%5E%7B%20%5Cfrac%7Bm%7D%7Bn%7D%7D)
Using these rules, we make all the indices, or exponents, look like
, and compare them to each other.
A.
B.
C.
![( \sqrt[3]{125} ) ^{7} =( 125^{ \frac{1}{3}} )^{7} = 125^{ \frac{1}{3}*7}= 125^{ \frac{7}{3}}](https://tex.z-dn.net/?f=%28%20%20%5Csqrt%5B3%5D%7B125%7D%20%29%20%5E%7B7%7D%20%3D%28%20125%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%7D%20%29%5E%7B7%7D%20%3D%20125%5E%7B%20%20%5Cfrac%7B1%7D%7B3%7D%2A7%7D%3D%20125%5E%7B%20%5Cfrac%7B7%7D%7B3%7D%7D%20%20%20)
D.
3x^2 - 4x
Im not quite sure what your asking, so I hope this helps!
