Answer:
![x =18.0](https://tex.z-dn.net/?f=x%20%3D18.0)
Step-by-step explanation:
To solve this problem use the Law of cosine.
The law of cosine says that:
![c^2 = a^2 + b^2 -2abcos(C)](https://tex.z-dn.net/?f=c%5E2%20%3D%20a%5E2%20%2B%20b%5E2%20-2abcos%28C%29)
In this case we have that:
![c = x\\\\a=30\\\\b=16\\\\C=30\°](https://tex.z-dn.net/?f=c%20%3D%20x%5C%5C%5C%5Ca%3D30%5C%5C%5C%5Cb%3D16%5C%5C%5C%5CC%3D30%5C%C2%B0)
Therefore
![x^2 = 30^2 + 16^2 -2(30)(16)cos(30\°)](https://tex.z-dn.net/?f=x%5E2%20%3D%2030%5E2%20%2B%2016%5E2%20-2%2830%29%2816%29cos%2830%5C%C2%B0%29)
![x = \sqrt{30^2 + 16^2 -2(30)(16)cos(30\°)}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B30%5E2%20%2B%2016%5E2%20-2%2830%29%2816%29cos%2830%5C%C2%B0%29%7D)
![x = \sqrt{1156 -831.38}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B1156%20-831.38%7D)
![x = \sqrt{324.62}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B324.62%7D)
![x =18.0](https://tex.z-dn.net/?f=x%20%3D18.0)
Answer:
i think it is A
Step-by-step explanation:
sorry if i am wrong...hope this helped, also can i have brainliest pls?
Answer:
The sum of the internal ángles = 360°
(3y+40)° and (3x-70°) are suplementary angles = 180°
then:
(3x-70) + (3y+40) + 120 + x = 360 ⇒ first eq.
(3y+40) + (3x-70) = 180 ⇒ second eq
development:
from the first eq.
3x + x + 3y = 360 + 70 - 40 - 120
4x + 3y = 430 - 160
4x + 3y = 270 ⇒ third eq.
3y = 270 - 4x
y = (270 - 4x) / 3 ⇒ fourth eq.
from the secon eq.:
3y + 3x = 180 + 70 - 40
3y + 3x = 250 - 40
3y + 3x = 210 ⇒ fifth eq.
multiply by -1 the fifth eq and sum with the third eq.
-3y - 3x = -210 ⇒ (fifth eq. *-1)
3y + 4x = 270
⇒ 0 + x = 60
x = 60°
from the fourth eq.
y = (270-4x)/3
y = (270-(4*60)) / 3
y = (270 - 240) / 3
y = 30/3
y = 10°
Probe:
from the first eq.
(3x-70) + (3y+40) + 120 + x = 360
3*60 - 70 + 3*10 + 40 + 120 + 60 = 360
180 - 70 + 30 + 40 + 120 + 60 = 360
180 + 30 + 40 + 120 + 60 - 70 = 360
430 - 70 = 360
Answer:
y = 10
Answer:
1. 8x+9
2. 3x + 11
3. 9x + 8
4. 11x+9
5. 6x
6. 10x+ 8
7. 7x+7
Step-by-step explanation: