What is one root of this equation? 2x^-4x+9=0
1 answer:
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Alright! Given that C(x) is Cost(Students) = 558, we can eliminate:
2. 558 students paid to attend the event.
5. The event generated $124 from student revenue.
Now, in order to find the cost per student we simply divide 124 on both sides:
124 cancels on the left and 558/124 is 4.5 or $4.50.
Since we just determined the cost of one student, we can eliminate 3.
To check if 124 students paid, we simply add the cost to the equation and check:
4.5(124) = 558
558 = 558 √
It checks out, so we determined that both 1 and 4 are correct.
2. 558 students paid to attend the event.
5. The event generated $124 from student revenue.
Now, in order to find the cost per student we simply divide 124 on both sides:
124 cancels on the left and 558/124 is 4.5 or $4.50.
Since we just determined the cost of one student, we can eliminate 3.
To check if 124 students paid, we simply add the cost to the equation and check:
4.5(124) = 558
558 = 558 √
It checks out, so we determined that both 1 and 4 are correct.
Your answers are
A = 35.7°
B = 67.6°
C = 76.7°
cosine law
![a^2 = b^2 + c^2 -2bc \cos A \\ -2bc \cos A = a^2 - b^2 - c^2 \\ \\ \cos A = \dfrac{a^2 - b^2 - c^2}{-2bc} \\ \\ A = \cos^{-1}\left[ \dfrac{a^2 - b^2 - c^2}{-2bc} \right] \\ \\ A = \cos^{-1}\left[ \dfrac{12^2 - 19^2 - 20^2}{-2(19)(20)} \right] \\ \\ A = 35.723697](https://tex.z-dn.net/?f=a%5E2%20%3D%20b%5E2%20%2B%20c%5E2%20-2bc%20%5Ccos%20A%20%5C%5C%0A-2bc%20%5Ccos%20A%20%3D%20a%5E2%20-%20b%5E2%20-%20c%5E2%20%5C%5C%20%5C%5C%0A%5Ccos%20A%20%3D%20%5Cdfrac%7Ba%5E2%20-%20b%5E2%20-%20c%5E2%7D%7B-2bc%7D%20%5C%5C%20%5C%5C%0AA%20%3D%20%5Ccos%5E%7B-1%7D%5Cleft%5B%20%5Cdfrac%7Ba%5E2%20-%20b%5E2%20-%20c%5E2%7D%7B-2bc%7D%20%5Cright%5D%20%5C%5C%20%5C%5C%0AA%20%3D%20%5Ccos%5E%7B-1%7D%5Cleft%5B%20%5Cdfrac%7B12%5E2%20-%2019%5E2%20-%2020%5E2%7D%7B-2%2819%29%2820%29%7D%20%5Cright%5D%20%20%5C%5C%20%5C%5C%0AA%20%3D%2035.723697)
A = 35.723697
sine law for the rest of the angles
![\displaystyle \frac{\sin B}{b} = \frac{\sin A}{a} \\ \\ \sin B = \frac{b \sin A}{a} \\ \\ B = \sin^{-1} \left[ \frac{b \sin A}{a} \right] \\ \\ B = \sin^{-1} \left[ \frac{19 \sin 35.723697 }{12} \right] \\ \\ B \approx 67.58886795](https://tex.z-dn.net/?f=%5Cdisplaystyle%0A%5Cfrac%7B%5Csin%20B%7D%7Bb%7D%20%3D%20%5Cfrac%7B%5Csin%20A%7D%7Ba%7D%20%5C%5C%20%5C%5C%0A%5Csin%20B%20%3D%20%5Cfrac%7Bb%20%5Csin%20A%7D%7Ba%7D%20%5C%5C%20%5C%5C%0AB%20%3D%20%5Csin%5E%7B-1%7D%20%5Cleft%5B%20%5Cfrac%7Bb%20%5Csin%20A%7D%7Ba%7D%20%20%5Cright%5D%20%5C%5C%20%5C%5C%0AB%20%3D%20%5Csin%5E%7B-1%7D%20%5Cleft%5B%20%5Cfrac%7B19%20%5Csin%2035.723697%20%7D%7B12%7D%20%20%5Cright%5D%20%20%5C%5C%20%5C%5C%0AB%20%5Capprox%2067.58886795)
B = 67.58886795
All angles in triangle sum to 180 so find C with that
A + B + C = 180
C = 180 - A - B
C = 180 - 35.723697 - 67.58886795
C = 76.7°
A = 35.7°
B = 67.6°
C = 76.7°
cosine law
A = 35.723697
sine law for the rest of the angles
B = 67.58886795
All angles in triangle sum to 180 so find C with that
A + B + C = 180
C = 180 - A - B
C = 180 - 35.723697 - 67.58886795
C = 76.7°