It should be a 100 degree angle as well.
9514 1404 393
Answer:
- 22.0
- 15.0
- 30.0°
- 137.0°
Step-by-step explanation:
These are all Law of Cosine problems. A generic expression for the length of side 'c' opposite angle C, which is defined by sides 'a' and 'b' is ...
c² = a² +b² -2ab·cos(C)
The square root of this gives the side length:
c = √(a² +b² -2ab·cos(C))
Rearranging the equation, we can obtain an expression for the angle C.
C = arccos((a² +b² -c²)/(2ab))
These two formulas are used to solve the offered problems.
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1) AC = √(13² +14² -2·13·14·cos(109°)) ≈ √483.506
AC ≈ 22.0
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2) BC = √(7² +10² -2·7·10·cos(123°)) ≈ √225.249
BC ≈ 15.0
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3) ∠B = arccos((24² +28² -14²)/(2·24·28)) = arccos(1164/1344)
∠B ≈ 30.0°
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4) ∠B = arccos((6² +9² -14²)/(2·6·9)) = arccos(-79/108)
∠B ≈ 137.0°
The graph you chose has a y intercept of 5. In other words, the point (0,5) is on that graph. However, if we plugged x = 0 into the function f(x), we get
f(x) = 3*5^x
f(0) = 3*5^0
f(0) = 3*1
f(0) = 3
So the point (0,3) is actually on the graph. The graph is increasing. So the answer is the upper left corner (instead of the upper right corner)
