Irregular Quadrilateral? I think?
264 because you can do 8 divided by 6 and get 1.3 repeating, than you would do 198 times 1.333333333333 and you get 264
Answer:
MAD = 0
Step-by-step explanation:
When I calculated this by hand I got 0. To find the MAD:
1. You first need to find the mean of the numbers.
4 + 5 + 8 + 8 + 10 = 35
35/5 = 7
the mean value = 7
2. Then you must find the absolute value of the difference between each data value and the mean: |data value – mean|
(4 - 7) = -3
(5 - 7) = -2
(8 - 7) = 1
(8 - 7) = 1
(10 - 7) = 3
3. Add the difference values up
(-3) + (-2) + 1 + 1 +3 = 0
4. Divide the sum of the absolute values of the differences by the number of data values.
0/5 = 0
Answer:
Adam: 52.26°
Step-by-step explanation:
Use Cosine Rule, 
, to find the angle of both players.
✔️Angle of Carlos:
Let the angle be C,
a = 50 ft
b = 40 ft
c = 24 ft
Plug in the values into the equation
C = 28.24° (nearest hundredth)
✔️Angle of Adam:
Let the angle be C,
a = 30 ft
b = 22 ft
c = 24 ft
Plug in the values into the equation
C = 52.26° (nearest hundredth)
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)
We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>
The vector product pq x pr gives a vector perpendicular to both pq and pr. This vector is the normal vector of a plane passing through all three points
pq x pr
=
i j k
-4 -2 -4
-3 5 1
=<-2+20,12+4,-20-6>
=<18,16,-26>
Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>
The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).
The equation of the required plane is therefore
Π : 9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π : 9x+8y-13z=24
Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.
