So... 2n>30 and 5n<100. So if you divide out 2 and 5, you get n>15 and n<20. Therefore 15<n<20
Answer:
1: 55 2: 40
Step-by-step explanation:
For the first problem, by the triangle angle theorm, the three angles in a triangle add up to 180, so
90+35+x=180
125 + x = 180
x = 55
Now, the angle you just found (x = 55) and the exterior angle are supplementary so they also add up to 180
55 + y = 180
y = 125
For the second problem, the two angles in a triangle non-adjacent to the exterior angle are equal to the exterior angle, so
40 + 2x = 3x
x = 40
Answer:
∠NMC = 50°
Step-by-step explanation:
The interpretation of the information given in the question can be seen in the attached images below.
In ΔABC;
∠ A + ∠ B + ∠ C = 180° (sum of angles in a triangle)
∠ A + 70° + 50° = 180°
∠ A = 180° - 70° - 50°
∠ A = 180° - 120°
∠ A = 60°
In ΔAMN ; the base angle are equal , let the base angles be x and y
So; x = y (base angle of an equilateral triangle)
Then;
x + x + 60° = 180°
2x + 60° = 180°
2x = 180° - 60°
2x = 120°
x = 120°/2
x = 60°
∴ x = 60° , y = 60°
In ΔBQC
∠a + ∠e + ∠b = 180°
50° + ∠e + 40° = 180°
∠e = 180° - 50° - 40°
∠e = 180° - 90°
∠e = 90°
At point Q , ∠e = ∠f = ∠g = ∠h = 90° (angles at a point)
∠i = 50° - 40° = 10°
In ΔNQC
∠f + ∠i + ∠j = 180°
90° + 10° + ∠j = 180°
∠j = 180° - 90°-10°
∠j = 180° - 100°
∠j = 80°
From line AC , at point N , ∠y + ∠c + ∠j = 180° (sum of angles on a straight line)
60° + ∠c + ∠80° = 180°
∠c = 180° - 60°-80°
∠c = 180° - 140°
∠c = 40°
Recall that :
At point Q , ∠e = ∠f = ∠g = ∠h = 90° (angles at a point)
Then In Δ NMC ;
∠d + ∠h + ∠c = 180° (sum of angles in a triangle)
∠d + 90° + 40° = 180°
∠d = 180° - 90° -40°
∠d = 180° - 130°
∠d = 50°
Therefore, ∠NMC = ∠d = 50°
Answer:
the total price is 64.89
Step-by-step explanation:
63 / 100 =
0.63 * 103 = 64.89