Answer:
2 roots in every case, not all are real.
Step-by-step explanation:
All of these equations are quadratic equations (degree 2). Every quadratic has two roots. They may be identical (looks like 1 root), and they may be complex (zero real roots), but there are always 2 of them.
a) the y-value of the vertex is negative and the parabola opens downward (leading coefficient -5), so there are no real zeros and both roots are complex.
b) each binomial factor contributes a root. Both roots are real.
c) the discriminant is positive, (3²-4·2·1=1), so both roots are real.
This question is incomplete because it lacks the diagram of the right angled triangle. Find attached to this answer the diagram of the right angle triangle.
Answer:
d-50
Step-by-step explanation:
Looking at the attached diagram, the only way to solve for this is the use of the trigonometric function. The trigonometric function to be used is the cosine function.
From the diagram, we are given
Hypotenuse = AB = 14
Adjacent = AC = 9
The measure of angle A to the nearest degree is calculated as:
cos θ = Adjacent / Hypothenuse
cos θ = 9/14
θ = cos -¹ (9/14) or arccos(9/14)
θ = 49.994799115°
To the nearest degree = 50°
Therefore,the measure of angle A to the nearest degree = 50°
2 ( x + 13) < 15
2x + 26 < 15
2x < -11
x < -5.5
Step-by-step explanation:
<h2>please mark me as brainlist please I </h2>
I think it would be $81 I’m not positive tho