The equation of the required plane can be obtained thus:
-4(x + 1) + 4(y + 3) + 3(z - 1) = 0
-4x - 4 + 4y + 12 + 3z - 3 = 0
4x - 4y - 3z = 5
Let x = 1, y = 2, then 4(1) - 4(2) - 3z = 5
z = (4 - 8 - 5)/3 = -9/3 = -3
Thus, point (1, 2, -3) is a point on the plane.
Let a = (a1, a2, a3) and b = (b1, b2, b3) be vectors parallel to the plane.
Then, -4a1 + 4a2 + 3a3 = 0 and -4b1 + 4b2 + 3b3 = 0
Let a1 = 2, a2 = -1, then a3 = (4(2) - 4(-1))/3 = (8 + 4)/3 = 12/3 = 4 and let b1 = -1 and b2 = 2, then b3 = (4(-1) - 4(2))/3 = (-4 - 8)/3 = -12/3 = -4
Thus a = (2, -1, 4) and b = (-1, 2, -4)
Therefore, the required parametric equation is r(s, t) = s(2, -1, 4) + t(-1, 2, -4) + (1, 2, -3) = (2s, -s, 4s) + (-t, 2t, -4t) + (1, 2, -3) = (2s - t + 1, -s + 2t + 2, 4s - 4t - 3)
Card P's balance increased by $3.43 more than Card Q's balance. The accumulated total on Card P over the 4 years is $1080.70 and the accumulated total of Card Q is $1,206.28. Based on the principal outlay however, Card P would have netted a higher interest over Card Q when the principal is subtracted from the accumulated value. (For eg. Card P accumulated value $1080.70 less Principal $726.19 equals $354.51).The interests over the 4 years period would be $354.51 and $351.08 respectively, hence Card P having an increase in balance of $3.43 over Card Q.
We know that 1 ft= 12 in.
9 ft* (12 in/ 1 ft)= 108 in.
15 inches to 9 feet
= 15 in: 9 ft
= 15 in: 108 in (because 9 ft= 108 in)
= 15/108
= (15/3) / (108/3)
= 5/36
The final answer is 5/36~
Simplifying
6y = 8 + -9 + 6y
Combine like terms: 8 + -9 = -1
6y = -1 + 6y
Add '-6y' to each side of the equation.
6y + -6y = -1 + 6y + -6y
Combine like terms: 6y + -6y = 0
0 = -1 + 6y + -6y
Combine like terms: 6y + -6y = 0
0 = -1 + 0
0 = -1
Solving
0 = -1
Couldn't find a variable to solve for.
This equation is invalid, the left and right sides are not equal, therefore there is no solution.
Consider : y² = 25
⇒ y = 
We know that : 5² = 25 and also (-5)² = 25
⇒ y = 
⇒ y = +5 or -5
So, The Answer is C
