In this problem, we could use the Angle Addition Postulate property. This property states that all interior angles within should sum of to the total angle. It is specifically stated that point H is interior of angle ∠JAK. Looking at the diagram attached in the picture, line segment AH is drawn in the middle. Therefore, the Angle Addition Postulate tells us that the sum of interior angles JAH and angle HAK, is equal to the total angle JAK.
∠JAK = ∠JAH + ∠HAK
∠JAK = (3x - 8) + (x+2)
But there is a missing information. Without knowing the total angle JAK, we can't solve for x. Consequently, we can't solve for the interior angles. So, let's just assume that ∠JAK = 45°. This is just for sample purposes.
45° = (3x - 8) + (x+2)
45 = 4x - 6
4x = 45+6
4x = 51
x = 12.75°
Therefore, the interior angles are equal to
∠JAH = 3(12.75) - 8 = 30.25°
∠HAK = 12.75 + 2 = 14.75°
Factor the polynomial.
−5(q+5)
plz mark me as brainliest if this helped :)
It is convenient to let a triangle solver figure this one.
The smallest angle is about 13.8°.
If you want to do it by hand, you can make use of the law of cosines.
a² = b² + c² - 2bc·cos(A)
Solving for the angle, you have
A = arccos((b²+c²-a²)/(2bc))
Substituting the given values gives you
A = arccos((23.8²+36.9²-14.9²)/(2·23.8·36.9))
A = arccos(1706.04/1756.44)
A ≈ arccos(0.9713056)
A ≈ 13.75879° ≈ 13.8°
So u have to rearrange to find x again
You look at the equation and u can see that if u take away 14 from both sides it will leave u with:
X/-3<-2
Now as x is over - 3, the opposite of divide is multiply, so you multiply both sides by - 3.
This leaves you with your answer:
x<6
So the number line which represents this is C.