Answer:
![\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-%5Cfrac%7B8%7D%7B%5Csqrt%7B117%7D%20%7D%20%5C%5C%5Cfrac%7B7%7D%7B%5Csqrt%7B117%7D%20%7D%5C%5C%5Cfrac%7B2%7D%7B%5Csqrt%7B117%7D%20%7D%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We are required to find a unit vector in the direction of:
![\left[\begin{array}{c}-8\\7\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-8%5C%5C7%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
Unit Vector, 
The Modulus of
=
Therefore, the unit vector of the matrix is given as:
![\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-%5Cfrac%7B8%7D%7B%5Csqrt%7B117%7D%20%7D%20%5C%5C%5Cfrac%7B7%7D%7B%5Csqrt%7B117%7D%20%7D%5C%5C%5Cfrac%7B2%7D%7B%5Csqrt%7B117%7D%20%7D%5Cend%7Barray%7D%5Cright%5D)
1. What is an equation of a line, in point-slope form, that passes through (1,-7) and has a slope of -2/3?
Point Slope form y − y1 = m(x − x1)
Y1: -7 x1:1 slope :-2/3
Y-(-7)=-2/3(x-1)
Y+7=-2/3(x-1)
2. What is the equation of a line, in point-slope form, that passes through (-2,-6) and had a slope of 1/3?
Y-(-6)=1/3(x-(-2))
Y+6=1/3(x+2)
3.What is an equation in point-slope form of the line that passes through the points (4,5) and (-3,-1)
SlopeM: =change in y/change in x
M= -1-5/-3-4
M= -6/-7
M=6/7
So now slope:6/7, point (4,5)
Y-y1=m(x-x1)
Equation in point slope
Y-5=6/7(x-4)
Answer:
(a) 7.5 seconds
(b) The horizontal distance the package travel during its descent is 1737.8 ft
Step-by-step explanation:
(a) The given function for the height of the object is s = -16·t² + v₀·t + s₀
The initial height of the object s₀ = 900 feet
The initial vertical velocity of the object v₀
= 0 m/s
The time it takes the package to strike the ground is found as follows;
0 = -16·t² + 0×t + 900
900 = 16·t²
t² = 900/16 = 62.25
t = √62.25 = 7.5 seconds
(b) Given that the horizontal velocity of the package is given as 158 miles/hour, we have
158 miles/hour = 231.7126 ft/s
The horizontal distance the package covers in the 7.5 second of vertical flight = 231.7126 ft/s × 7.5 s = 1737.8445 feet = 1737.8 ft to one decimal place
The horizontal distance the package travel during its descent = 1737.8 ft.
For 61 you can change it to √25/√49. √25=5 and √49=7 so it would simplify to 5/7.
The number of data values between the lower quartile and the median is less than the number of data values between the upper quartile and the median.
Hope this helps!