Answer:
C
Step-by-step explanation:
400-160=240
240/4=60
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
Answer:
x=2
Step-by-step explanation:
use distributive property: x^2+2x-1=4
remove the -1 by adding 1 on both sides: x^2+2x=5
use the guadratic fromula: ax²+bx+c=0
plug in the equation x^2+2x-5=0
a,b and c are the coefficients to plug into the formula
a=1, b=2, c=-1
Finally, you will find that x=2
The equation for a hyperbola is (x-h)/a - (y-k)/b = 1
Or (y-k)/a - (x-h)/b = 1
h represents the x value of the coordinate
k value represents the y value of the coordinate
together they represent a point, which is the center
So (h,k) is (x,y)
The asymptote is y-k = +/- b/a (x-h)
The transverse is the line that goes through the hyperbola.
Answer:
Card 7
Step-by-step explanation:
2+5+7+8+9 =31
Multiples of 6: 6, 12, 18, 24, 30
31-9=22, No
31-8=23, No
31-5=26, No
31-2=29, No
31-7=24, Yes
The only way to get a multiple of 6 is to subtract card 7, so card 7 is the answer.
