So it (-6-22)+(321) ? Before I solve this problem?
Answer:
The answer to your question is 50 inches
Step-by-step explanation:
Data
Area = 200 in²
diagonal 1 = 8 in
diagonal 2 = ?
Formula
Area = (diagonal 1 x diagonal 2) / 2
Solve for diagonal 2
diagonal 2 = 2 x Area 2 / diagonal 1
-Substitution
diagonal 2 = 2 x 200 / 8
-Simplification
diagonal 2 = 400 / 8
-Result
diagonal 2 = 50 in
Your sequence appears to be geometric with a common ratio of 2. It can be described by
a(n) = (-2 2/3)·2^(n-1)
_____
This can be written in a number of other forms, including
a(n) = (-8/3)·2^(n-1)
a(n) = (-1/3)·2^(n+2)
a(n) = (-4/3)·2^n
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:
C!
Step-by-step explanation:
