I don't think that's possible.
The expression in company B represents that is is in arithmetic progression where first term is 42000 and common difference is 1800 . So we have to use the formula of sum of n terms which is
Where a is the first term, n is the nth term, d is the common difference
On substituting there values,we will get
= 15(84000+52200) = 15*136200 =2043000
And for company A, it is
Difference between them =2043000-2002500= 40500
So the correct option is the second option .
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 = 154^3+10^2+12+15
Let a₁ , a₂ , a₃ , a₄ ,... .be a given sequence.
The common ratio of this sequence is the following:
a₂/a₁ = a₃/a₂ = a₄/a₃ = r
Example: 5, 25, 125, 625, ...The common ration is:
25/5 = 125/25 = 625/125 = 5. So r=5 is the common ratio
