<h2>Answer:
</h2>
The length of VW is 23°
<h2>Step-by-step explanation:
</h2><h3>Known :
</h3>
<h3>Asked :
</h3>
VW = ?
<h3>Solution :
</h3>
Since the UV line and UX line has the same length, so the VW line should have the same length as the WX line.
VW = WX
23 = 23
<h3>Conclusion :
</h3>
The length of VW is 23°
Answer:
<h2>
x=1</h2>
Step-by-step explanation:
|20x|=|4x+16|
or, 20x-4x=16
or, 16x=16
Hence, x=1
Looking at the angles of the triangles
N=S
M=R
O=T
so MON would be RTS
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.
