Find the powers 
$a^{2}=5+2 \sqrt{6}$
$a^{3}=11 \sqrt{2}+9 \sqrt{3}$
The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.
Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$
so fits with the other answers.
Step-by-step explanation:
<YOZ = 60°
When two lines intersect each other, then four angles are formed. The angles that are opposite to each other at the intersection point are known as <em>vertically</em><em> </em><em>opposite</em><em> </em><em>angles</em><em>.</em><em> </em>VERTICALLY OPPOSITE ANGLES ARE ALWAYS EQUAL.
If < VOX = 90⁰, then <VOW = 60⁰.
<VOW and <YOZ are vertically opposite angles.
Answer:
∠TAR = 35°
Step-by-step explanation:
∠TAR = ∠CAR - ∠CAT = (9x + 7) - 5x = 4x + 7
AT bisect ∠CAR ∠TAR = ∠CAT
4x + 7 = 5x
x = 7
∠TAR = 4 x 7 + 7 = 35
If the point is at (-3, -4), it is located in the third quadrant. You can find the tangent of an angle by finding the ratio of the length of the opposite side to the length of the adjacent side. In this case, the opposite side of the angle is the vertical line with a length of 4 units, while the adjacent side of the angle is the horizontal line with a length of 3 units. <em>Thus, tan θ = 4/3.</em>
