Answer:
3x^2 + 3xz + 2y^2 - 2yz - xy
Step-by-step explanation:
(1) multiply it to get
3x^2 - 3xy + 3xz and 2yx + 2y^2 - 2yz
add similar variables (-3xy + 2yx)
( the order of the variables doesn't matter xy or yx is the same)
we get -xy
and then after that simplifying we get 3x^2 + 3xz + 2y^2 - 2yz - xy
I think the answer is B but I don't know for sure.
I think that the formula for finding the volume of a right pyramid is V= 1/3Bh (Volume equals one third base times height).
To find the base we need to take the Length and multiply it by the Width (L x W). Using the information given, we know the the base is a square and that one side of the square is 6. All sides of a square are equal so the length and the with are both 6. Now we multiply 6 (length) by 6 (width) and we get 36. 36 is our base.
We now have to find the height or altitude. In the given information it says that the altitude it equal to the side of the base. The side of the base is 6, so 6 is our altitude.
Now we plug the numbers into our formula.
V= 1/3(36)(6)
Multiply 36 by 6 and we get the answer 216, but we still need to divide it by 1/3. 216 divided by 1/3 equals 72.
Our answer is 72m3!!!!!
Hope this helps and let me know if I'm wrong.
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.
To solve the two sets of equations simultaneously, subtract one equation from the other to obtain
3x^5 + 2x^2 - 10x + 4 - (4x^4 + 6x^3 - 11) = 0
3x^5 - 4x^4 - 6x^3 + 2x^2 - 10x - 7 = 0
This is a polynomial of degree 5 to be solved for zeros.
A graphing calculator will yield 3 real zeros (verifiable by Descartes Rule of Signs).