L=7+w
P=(12x+14)=2(6x+7)
l+w= P/2 = 2(6x+7)/2= 6x+7
l+w=6x+7
2l = 6x+7
l = (6x+7)/2
l=7+w
-> w= l-7
= [(6x+7)/2] -7
= (6x+7-14)/2
= (6x-7)/2
Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
114.93 because when you add them you get that answer
The picture in the attached figure
[surface area of the composite solid]=1*(4*6)+2*(4*4)+2*(4*6)+2*(4*√13/2)+2*(6*2√2/2)
[surface area of the composite solid]=24+32+48+4√13+12√2
[<span>surface area of the composite solid]=135.39 yd</span>²
the answer is135.39 yd²