Answer:
9. 66°
10. 44°
11. 
12. 
13. 27.3
14. 33.9
15. 22°
16. 24°
Step-by-step explanation:
9. Add 120 + 80 (equals 200) and subtract that from 360 (Because all angles in a quadrilteral add to 360°), this equals 160. Plug the same number in for both variables in the two other angle equations until the two angles add to 160. For shown work on #9, write:
120 + 80 = 200
360 - 200 = 160
12(5) + 6 = 66°
19(5) - 1 = 94°
94 + 66 = 160
10. Because the two sides are marked as congruent, the two angles are as well. This means the unlabeled angle is also 68°. The interior angles of a triangle always add to 180°, so add 68+68 (equals 136) and subtract that from 180, this equals 44. For shown work on #10, write:
68 x 2 = 136
180 - 136 = 44
11. Use the Pythagorean theorem (a² + b² = c²) (Make sure to plug in the hypotenuse for c). Solve the equation. For shown work on #10, write:
a² + b² = c²
a² + 6² = 8²
a² + 36 = 64
a² = 28
a = 
a = 
12. (Same steps as #11) Use the Pythagorean theorem (a² + b² = c²) (Make sure to plug in the hypotenuse for c). Solve the equation. For shown work on #11, write:
a² + b² = c²
a² + 2² = 4²
a² + 4 = 16
a² = 12
a = 
a = 
13. Use SOH CAH TOA and solve with a scientific calculator. For shown work on #13, write:
Sin(47°) = 
x = 27.3
14. Use SOH CAH TOA and solve with a scientific calculator. For shown work on #14, write:
Tan(62°) = 
x = 33.9
15. Use SOH CAH TOA and solve with a scientific calculator. For shown work on #15, write:
cos(θ) = 52/56
θ = cos^-1 (0.93)
θ = 22°
16. (Same steps as #15) Use SOH CAH TOA and solve with a scientific calculator. For shown work on #16, write:
sin(θ) = 4/10
θ = sin^-1 (0.4)
θ = 24°
Good luck!!
For two shots Pedro has four outcomes:
<u>1 shot | 2 shot</u>
score | score
score | not score
not score | score
not score | not score
The probability Pedro's shot will score in a lacrosse game is 0.30 and the probability his shot will not score in a lacrosse game is 1-0.30=0.70. So you can count probabilities for all cases:
1. 0.3·0.3=0.09;
2. 0.3·0.7=0.21;
3. 0.7·0.3=0.21;
4. 0.7·0.7=0.49.
In total 0.09+0.21+0.21+0.49=1. The first outcomes is that what you need.
Answer: 0.09.
Answer:
I'm not SUPER sure but, it only can exist where there is a region in a planetary system where liquid water might exist on the surface of a planet. That puts it in the "habitable zone" which is where it can exist.
The answer is 24
you do 6+6+6+6