Answer:
C
Step-by-step explanation:
Using the rule of radicals/ exponents
⇔ ![\sqrt[n]{a^{m} }](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5E%7Bm%7D%20%7D)
Given
= ![\sqrt[3]{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5%7D)
→ C
47x-12
by step
36x -12 +11
Answer:
2
Step-by-step explanation:
The height formula given is:
h = -16t^2 + 70
That means the object will be initially (t=0) at the height 70 feet, from where it will be dropped.
If we want to know the time when the object will be at height 6 feet, we just need to use h=6 in the equation, and then calculate the value of t:
6 = -16t^2 + 70
16t^2 = 64
t^2 =4
t = 2 s
So, it will take 2 seconds for the object to be 6 feet above the valley floor.
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
5(a + 3b) = 5(9 + 3(2)) = 5(9 + 6) = 5(15) = 75
