Q1. The answer is x = 1, y = 1, z = 0
<span>(i) -2x+2y+3z=0
</span><span>(ii) -2x-y+z=-3
</span>(iii) <span>2x+3y+3z=5
</span><span>_________
Sum up the first and the third equation:
</span>(i) -2x+2y+3z=0
(iii) 2x+3y+3z=5
_________
5y + 6z = 5
Sum up the second and the third equation:
(ii) -2x-y+z=-3
(iii) 2x+3y+3z=5
_________
2y + 4z = 2
(iv) 5y + 6z = 5
(v) 2y + 4z = 2
________
Divide the fifth equation by 2
(iv) 5y + 6z = 5
(v) y + 2z = 1
________
Multiple the second equation by -3 and sum the equation
(iv) 5y + 6z = 5
(v) -3y - 6z = -3
________
2y = 2
y = 2/2 = 1
y + 2z = 1
1 + 2z = 1
2z = 1 - 1
2z = 0
z = 0
-2x-y+z=-3
-2x - 1 + 0 = -3
-2x = -3 + 1
-2x = -2
x = -2/-2 = 1
Q2. The answer is x = -37, y = -84, z = -35
<span>(i) x-y-z=-8
(ii) -4x+4y+5z=7
(iii) 2x+2z=4
______
</span>Divide the third equation by 2 and rewrite z in the term of x:
(iii) x+z=2
z = 2 - x
______
Substitute z from the third equation and express y in the term of x:
<span>x-y-(2-x)=-8
x - y - 2 + x = 8
2x - y = 10
y = 2x - 10
______
Substitute z from the third equation and y from the first equation into the second equation:
</span><span>-4x + 4y + 5z = 7
-4x + 4(2x - 10) + 5(2 - x) = 7
-4x + 8x - 40 + 10 - 5x = 7
-x -30 = 7
-x = 30 + 7
x = -37
y = 2x - 10 = 2*(-37) - 10 = -74 - 10 = -84
z = 2 - x = 2 - 37 = -35</span>
D) 4.5 Standard Deviations Above the Mean
All you have to do is add .2 to 6.6 until you reach 7.5
6.8
7.0
7.2
7.4
Then you only need half of .2 to reach 7.5, so it's 4.5 standard deviations above the mean
Answer:
-6
Step-by-step explanation:
Answer:
x>9 this is the correct answer
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π and cos A = cos B · cos C
scratchwork:
A + B + C = π
A = π - (B + C)
cos A = cos [π - (B + C)] Apply cos
= - cos (B + C) Simplify
= -(cos B · cos C - sin B · sin C) Sum Identity
= sin B · sin C - cos B · cos C Simplify
cos B · cos C = sin B · sin C - cos B · cos C Substitution
2cos B · cos C = sin B · sin C Addition
Division
2 = tan B · tan C
<u>Proof LHS → RHS</u>
Given: A + B + C = π
Subtraction: A = π - (B + C)
Apply tan: tan A = tan(π - (B + C))
Simplify: = - tan (B + C)
Substitution: = -(tan B + tan C)/(1 - 2)
Simplify: = -(tan B + tan C)/-1
= tan B + tan C
LHS = RHS: tan B + tan C = tan B + tan C 
