If -3 + i is a root of the polynomial function f(x), -3 - i must also be a root of f(x)
<h3>How to determine the true statement?</h3>
The root of the polynomial function is given as:
-3 + i
The above root is a complex root.
If a polynomial has a complex root, then the conjugate of the root is also a root of the function
The conjugate of -3 + i is -3 - i
Hence, if -3 + i is a root of the polynomial function f(x), -3 - i must also be a root of f(x)
Read more about polynomial functions at:
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Answer:
x=1
Step-by-step explanation:
7x - 20 = 2x - 3(3x + 2)
Distribute
7x -20 = 2x -9x -6
Combine like terms
7x-20 =-7x -6
Add 7x to each side
7x+7x-20 =-7x+7x -6
14x -20 = -6
Add 20 to each side
14x-20+20 =-6+20
14x = 14
Divide by 14
14x/14=14/14
x=1
Answer:
1. Yes the parentheses are necessary. To find a fourth of her regular hours you must find the total amount she works during her regular hours.
2. 6
Step-by-step explanation:
(For the second question)
4+8= 12
12 × ½ = 12 ÷ 2
12÷2 = 6
If A is QIV, then 3π/2 ≤A≤2π;
we have to find out in what quadrant is A/2
(3π/2)/2≤A/2≤(2π)/2 ⇒ 3π/4≤A/2≤π
We can see that A/2 will be in QIII; therefore the sec (A/2) will be negative (-).
1) we have to calculate cos (A/2)
Cos (A/2)=⁺₋√[(1+cos A/2)/2]
We choose this formula: Cos (A/2)= -√[(1+cos A/2)/2], because sec A/2 is in quadrant Q III, and the secant (sec A/2=1/cos A/2) in this quadrant is negative.
Cos (A/2)=-√[(1+cos A)/2]=-√[(1+(1/2)]/2=-√(3/4)=-(√3)/2.
2) we compute the sec (A/2)
Data:
cos (A/2)=-(√3)/2
sec (A/2)=1/cos (A/2)
sec (A/2)=1/(-(√3)/2)=-2/√3=-(2√3)/3
Answer: sec (A/2)=-(2√3)/3
You would take -9 - -2 and then add 4 and multiply
