Answer:
0
1
Step-by-step explanation:
First question:
You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.




The sine function can never equal 2, so there is no triangle in this case.
Answer: no triangle
Second question:
You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.





One triangle exists for sure. Now we see if there is a second one.
Now we look at the supplement of angle C.
m<C = 52.5°
supplement of angle C: m<C' = 180° - 52.5° = 127.5°
We add the measures of angles B and the supplement of angle C:
m<B + m<C' = 63° + 127.5° = 190.5°
Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.
Answer: one triangle
Answer:
- 3x² + 5x + 1 = 0 simplified
- x = (-5 ±
)/6 roots
Step-by-step explanation:
<u>Given quadratic formula</u>
<u>Standard form</u>
<u>Simplifying</u>
- 3x² + 5x - 5 + 6 = 0
- 3x² + 5x + 1 = 0
<u>Solving</u>
- x = (-5 ±
)/(2*3) - x = (-5 ±
)/6
Answer:
it is b for sure sokays uduhdhd
Answer:
D.24
Pythagorean states:
hypotenuse^2 = side1^2 + side2^2
30^2 = 18^2 + side2^2
side2^2 = 30^2 - 18^2
side2^2 = 900 - 324
side2^2 = 576
side2 = sqr root 576
side2 = 24
Source: http://www.1728.org/pythgorn.htm
Step-by-step explanation: