Given:
P: (2,0,5)
L: (0,6,4)+t(7,-1,5)
and required plane, Π , passes through P and perpendicular to L.
The direction vector of L is V=<7,-1,5>.
For Π to be perpendicular to V, Π has V as the normal vector.
The equation of a plane with normal vector <7,-1,5> passing through a given point P(xp,yp,zp) is
7(x-xp)-1(y-yp)+5(z-zp)=0
Thus the equation of plane Π passing through P(2,0,5) is
7(x-2)-y+5(z-5)=0
or alternatively,
7x-y+5z = 14+25
7x-y+5z = 39
The answer is 813.7 your welcome
Answer:
Step-by-step explanation:
<u>Option a,c,e,f,i,j are quadratic equations. (Because its degree is 2)</u>
<u>Option b,d,g are linear equations. (Because its degree is 1)</u>
<u>Just option h is not quadratics nor linear.</u>
<u />
<h3><u>If you need to ask any questions, please let me know.</u></h3>
So factor
use reverse distributiver property
ab+ac=a(b+c) or
ab-ac=a(b-c)
so we need to find a
find the common factors of the variables
4=2 times 2
8=2 times 2 times 2
common is 2 times 2=4
so 4 is a factor of both
so
4x-8=(4)x-(4)2=(4)(x-2)
4(x-2) is the answer