Answer:
t = 9.08 sec
Explanation :
This requires the use of the kinematic equations.
u = initial velocity = 21.8 m/s
v = final velocity = 0
t = time = what we're trying to find.
So we use the equation that relates d, u, v and t. This is
d = distance = 99m
d = 
and since v = 0 this reduces to d = 
which rearranges to t = 
t = 
=9.08 seconds
t = 9.08 sec
Answer: 4 feet.
Step-by-step explanation:
Their is one rectangular prison divided into two.
The volume of the first one is
V1= L X H X W=80.
Volume of the second one is
V2= L X H X W
V2= 6 X H X 5
V2 = 30H
Now from the question, the total volume is 200cubic feet which means V1 + V2 = 200, recall that V1=80 and V2=30H, so we say;
80+30H=200, collecting the like term we have, 30H=200-80.
30H=120, divide both side by 30
30H/30=120/30
H= 4feet.
Height is 4feet.
Answer:
There doesn't seem to be numbers to answer that, do you have the question right? If it was a difference of 18, for example, the numbers would be 64 and 36.
Step-by-step explanation:
Find 2 numbers that added together, equal 10.
91
82
73
64
55
46
37
28
19
Then subtract them by their opposites. Find the numbers with the difference of 12.
91 - 19 = 72
82 - 28 = 54
73 - 37 = 36
64 - 46 = 18
55 - 55 = 0
46 - 64 = -18
37 - 73 = -36
28 - 82 = -54
19 - 91 = -72
There is no numbers with the difference of 12.
If it was a difference of 18, for example, the numbers would be 64 and 36.
let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD
so we're really looking for the equation of a line whose slope is -1/3 and passes through point A
![(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20-%5Ccfrac%7B1%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B3%7D%3D%5Cstackrel%7Bm%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7D%5Bx-%5Cstackrel%7Bx_1%7D%7B%28-1%29%7D%5D%5Cimplies%20y-3%3D-%5Ccfrac%7B1%7D%7B3%7D%28x%2B1%29%20%5C%5C%5C%5C%5C%5C%20y-3%3D-%5Ccfrac%7B1%7D%7B3%7Dx-%5Ccfrac%7B1%7D%7B3%7D%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx-%5Ccfrac%7B1%7D%7B3%7D%2B3%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx%2B%5Ccfrac%7B8%7D%7B3%7D)
Answer:
y=(x+3)^2 has a vertex at (-3,0), while y=x^2 + 3 has a vertex at (0,3)
Mark me brainliest hope this helps
Step-by-step explanation:
