X^3 = 216
by taking cubic root for both sides
![\sqrt[3]{x^3} = \sqrt[3]{216}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%5E3%7D%20%3D%20%20%5Csqrt%5B3%5D%7B216%7D%20)
x = 6
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:
someone is adding dont worrie
Step-by-step explanation:
Answer:
the answer is b, 12
Step-by-step explanation:
16-4=12
So basically sam split the 1000 to 2 groups so we have 1000 ÷ 2
now the first group divided the 500 (since 1000÷ 2 is 500) among five children so it would be 500 ÷ 5 which equals 100 per child and the second group divided their 500 between to boys, so you would get 500÷2 which is 250
so for group one it would be: 1000÷2÷5 which equals 100 per child and the second group is 1000÷2÷2 which equals 250 for each brother
