The product of a variable and a number should be a variable and a number. I think A, b, and d are the answer because they all comprise of a variable (c, t, j) and a number (11, 1/2, 4). C only provides two numbers and no variable
Trigonometric Identities.
To solve this problem, we need to keep in mind the following:
* The tangent function is negative in the quadrant II
* The cosine (and therefore the secant) function is negative in the quadrant II
* The tangent and the secant of any angle are related by the equation:

We are given:
![\text{tan}\theta=-\frac{\sqrt[]{14}}{4}](https://tex.z-dn.net/?f=%5Ctext%7Btan%7D%5Ctheta%3D-%5Cfrac%7B%5Csqrt%5B%5D%7B14%7D%7D%7B4%7D)
And θ lies in the quadrant Ii.
Substituting in the identity:
![\begin{gathered} \sec ^2\theta=(-\frac{\sqrt[]{14}}{4})^2+1 \\ \text{Operating:} \\ \sec ^2\theta=\frac{14}{16}+1 \\ \sec ^2\theta=\frac{14+16}{16} \\ \sec ^2\theta=\frac{30}{16} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Csec%20%5E2%5Ctheta%3D%28-%5Cfrac%7B%5Csqrt%5B%5D%7B14%7D%7D%7B4%7D%29%5E2%2B1%20%5C%5C%20%5Ctext%7BOperating%3A%7D%20%5C%5C%20%5Csec%20%5E2%5Ctheta%3D%5Cfrac%7B14%7D%7B16%7D%2B1%20%5C%5C%20%5Csec%20%5E2%5Ctheta%3D%5Cfrac%7B14%2B16%7D%7B16%7D%20%5C%5C%20%5Csec%20%5E2%5Ctheta%3D%5Cfrac%7B30%7D%7B16%7D%20%5Cend%7Bgathered%7D)
Taking the square root and writing the negative sign for the secant:
We know that
<span>(16x6 − 12x4 + 4x2) / 4x2
is equal to
[16x6/4x2]+[-12x4/4x2]+[4x2/4x2]
=[4x4]+[-3x2]+[1]
=4x4-3x2+1
the answer is
</span>4x4-3x2+1<span>
</span>
Answer:
c
Step-by-step explanation:
got you g
Answer:

Step-by-step explanation:
1) In this question we've been given "a", the leading coefficient. and two roots:

2) There's a theorem, called the Irrational Theorem Root that states:
If one root is in this form
then its conjugate
. is also a root of this polynomial.
Therefore

3) So, applying this Theorem we can rewrite the equation, by factoring. Remembering that x is the root. Since the question wants it in this expanded form then:
