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:
9.72
Step-by-step explanation:
s1 = 10.6383 ; s2 = 5.21289
x1 = 147.583 ; x2 = 136.417
n1 = 12 ; n2 = 12
df1 = n1 - 1 = 12 - 1 = 11
df2 = n2 - 1 = 12 - 1 = 11
The test statistic :
(x1 - x2) / sqrt[(sp²/n1 + sp²/n2)]
Pooled variance = Sp² = (df1*s1² + df2*s2²) ÷ (n1 + n2 - 2)
Sp² = ((11*10.6383) + (11*5.21289)) / 22 = 7.926
Test statistic, T* :
(147.583 - 136.417) / √(7.926 * (1/12 + 1/12))
11.166 / √(7.926 * (1/6)
11.166 / √1.321
11.166 / 1.1493476
T* = 9.7150766
Test statistic = 9.72
Here are both in fraction form.
i) 7/100
ii) 6/9
Y=2-3(x-7)
You just add +2 to each side and the positive and negative cancel out and it switches it over
Answer:
1
Step-by-step explanation:
(195 - 85)/2 = 54x+1 -->
110/2 = 54x + 1 -->
55 = 54x + 1 -->
54 = 54x -->
1 = x
