Answer:
<em>Alex sold 330 white chocolate bars</em>
Step-by-step explanation:
let the milk chocolate candy be x and the white chocolate candy be y. then
then 2x+3y=990----1)
x+y=330-----2)
from eq 2) ⇒ x=330-y---------3)
Putting eq 3 in eq 1
2(330-y) +3y=990
660-2y+3y=990
y=990-660=330-----------4)
Putting eq4 in eq1
2x+3(330)=990
2x+990=990
2x=990-990
2x=0
x=0---------------------------5)
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.
x^3 - 3x^2 - 18x
Using the FOIL method, I arrived at my solution!
For the first one it’s b and I’m sorry but I’m not sure about the second
