1. A.Use law of cosines. cosA=(b^2+c^2-a^2)/(2bc) because A is the included angle between b and c. Plug in A=50 degrees, b=13, c=6. cos50=(13^2+6^2-a^2)/(2*13*6), a^2=104.7,a=10.2 approximately, so choose A.
2. A.Use law of cosines again. cosC=(a^2+b^2-c^2)/(2ab). C=95 degrees, a=12, b=22. Plug in, cos95=(12^2+22^2-c^2)/(2*12*22), solve the equation, c^2=674, c=26 approximately. The use law of sines to solve for angle A (also works for B), a/sinA=c/sinC, 12/sinA=26/sin95, sinA=0.46, A=arcsin(0.46)=27.6. Choose A.
3. Answer is A. Area=1/2bc*sinA, since A is the included angle between b and c. Plug in b=30, c=14, A=50 degrees, area=1/2*14*30*sin50=160. 87, so the answer is A.
4. D. As long as the sum of any two sides of the triangle is bigger than the third, the triangle exists. 240+121>263, 240+263>121, 263+121>240, so it exists. To use Heron's formula, first find the semiperimeter, (240+263+121)/2=312. A=\sqrt(312*(312-240)*(312-263)*(312-121))=14499.7 approximately, so choose D.
5. 300. The included angle between the two paths is C=49.17+90=139.17 degrees. The lengths of the two paths are a=150, b=170. c is the distance we want. Use the law of cosines, cosC=(a^2+b^2-c^2)/(2ab). Plug in, c^2=89989, c=300 approximately.
Rise=16, run=7. Slope is rise/run, and therefore is 16/7. The run is considered to be the x coordinate. Since the x coordinate on the point is 21, we know it has moved 3 exact points away from the origin (21/7=3). We can use this movement of 3 exact points to determine the y coordinate as well. Since the rise (y) is 16 for every exact point on a graph, we know the graph has risen 48 units (16x3=48). So, the point ends up being (21,48). The rate of change is 16cm per 7 minutes, or 2.28cm/minute. The equation is Y=(16/7)X (no y intercept because the graph starts at the origin). The equation gives you the y value of 48 when x is equal to 21.
First you need to know the prime numbers from 1-10. They are 2, 3, 5, 7, 1st draw: P(prime) = 4/10 2nd draw: P(prime) = 3/9 Multiply these numbers 12/90 = 2/15
