For perpendicular lines, the slopes are negative reciprocals of one another. Therefore, the slope of QC = 3/2
Using this into the formula of slope:
3/2 = (y + 6) / (-1 + 3)
y = -9
If the slope of CD is 2/3, then y works out to be -3.
Answer:
x = -15/2
Step-by-step explanation:
For this problem, we will simply use equation properties to solve for x.
2x - 5 = -20
2x - 5 + 5 = -20 + 5
2x = -15 ( Add positive 5 to both sides )
2x * (1/2) = -15 * (1/2)
x = -15/2 ( Multiply both sides by 1/2)
Hence, the solution to x is -15 / 2.
Cheers.
Let p = number of people that cast their vote.
p = 0.35(15, 000)
p = 5,250
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.
Answer:
2/10
Step-by-step explanation:
0.20 = 20/100
20/100 = 2/10
