PEMDAS
(9)(5) + 8
45 + 8
53
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.
8 is to 64 as 2 is to 16
x=16
Answer:
its 2
Step-by-step explanation:
rise/run. it goes up one and over .5 each time, so 1÷0.5=2.
Answer: x = 3, y = 1, z = 2
<u>Step-by-step explanation:</u>
EQ 1: x - y - z = 0
EQ 3:<u> -x + 2y + z = 1 </u>
y = 1
EQ 2: 2x - 3y + 2z = 7 → 1(2x - 3y + 2z = 7) → 2x - 3y + 2z = 7
EQ 3: -x + 2y + z = 1 → -2( -x + 2y + z = 1) → <u>-2x + 4y + 2z = 2</u>
y + 4z = 9
y = 1 ⇒ 1 + 4z = 9
4z = 8
z = 2
Input y = 1 and z = 2 into one of the equations to solve for x:
EQ 1: x - y - z = 0
x - (1) - (2) = 0
x - 3 = 0
x = 3
Check:
EQ 2: 2x - 3y + 2z = 7
2(3) - 3(1) + 2(2) = 7
6 - 3 + 4 = 7
3 + 4 = 7
7 = 7 