Answer:
Option D
Step-by-step explanation:
Option A
x² - 3x² + 2x² = 0
Therefore, the given polynomial has no roots.
Option B
x² + 2x + 8
By quadratic formula,
x = 
= 
Therefore, roots are imaginary and for a polynomial with degree 1 or more than 1, both the roots can't be imaginary.
So the polynomial can't have exactly two roots.
Option C
99x³ - 33x + 1
Since the polynomial is of degree 3, so it will have three roots.
Therefore, the polynomial will not have exactly 2 roots.
Option D
√2x - 3x² + 7√2
By quadratic formula,
x = 
x = 
There are exactly two real roots.
Therefore Option D is the answer.
Option E
4x + 11x - 111
Since this polynomial is of a degree 1.
There will be only one root.
Answer:
Cos(π/2 - θ) = A
Step-by-step explanation:
We know that:
Sin(θ) = A
And we want to find the value of Cos(π/2 - θ)
Here we can use the cosine relationship:
Cos(a - b) = Cos(a)*Cos(b) + Sin(A)*Sin(B)
Then:
Cos(π/2 - θ) = Cos(π/2)*Cos(θ ) + Sin(π/2)*Sin(θ)
We know that:
Cos(π/2) = 0
Sin(π/2) = 1
Then:
Cos(π/2 - θ) = Cos(π/2)*Cos(θ ) + Sin(π/2)*Sin(θ) = 0*Cos(θ ) + 1*Sin(θ)
= 1*Sin(θ) = A
Cos(π/2 - θ) = A
I believe it’s 6, so E, but I also believe I’m completely wrong-
Answer:
x=3
Step-by-step explanation:
Secant-Secant Formula:
(whole secant) x (external part) = (whole secant) x (external part)
9*4 = 12*x
36 = 12x
Divide each side by 12
36/12 = 12x/12
3 =x