The least number of buses needed to carry 710 passengers is 5.
Option A)5 is the correct answer.
<h3>What is the least number of buses needed to carry 710 passengers? </h3>
Given that;
- Number of passengers n = 710
- Least number of buses need B = ?
To get the least number of buses, we say;
Number of buses B = 710 ÷ 150
Number of buses B = 4.7 ≈ 5
The least number of buses needed to carry 710 passengers is 5.
Option A)5 is the correct answer.
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Answer:
The correct answer to the question is;
1 < x < <u>4</u>
Step-by-step explanation:
The give parameters are;
ΔABC is an isosceles triangle, with the vertex angle = 35° + 20° = 55°
Therefore, the two base angles are 62.5° each by definition of isosceles triangle
Given that the angle subtended by 4·x - 4 which is 25° is lesser than the angle subtended by the 12 unit side length which is 35° we have;
4·x - 4 < 12
4·x < 12 + 4
∴ 4·x < 16
x < 16/4
x < 4
Therefore, the range is 1 < x < 4
Answer:
3.48%
Step-by-step explanation:
Probability = Area of hole / Area of board
= 3.14 x 2 x 2 / 12 x 30
= 12.56 / 360
= 0.0348
= 3.48%
#1)
A) b = 10.57
B) a = 22.66
#2)
A) a = 1.35 (across from the 15° angle)
∠C = 50.07° (the angle at the top of the triangle)
∠B = 114.93°
B) ∠A = 83°
b = 10.77 (across from angle B)
a = 15.11 (across from angle A)
Explanation
#1)
A) Since b is across from the 25° angle and we have the hypotenuse, we have the information for the sine ratio (opposite/hypotenuse):
sin 25 = b/25
Multiply both sides by 25:
25*sin 25 = (b/25)*25
25*sin 25 = b
10.57 = b
B) We will first use the cosine ratio. Side a is the side adjacent to the angle and we have the hypotenuse, and the cosine ratio is adjacent/hypotenuse:
cos 25 = a/25
Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a
Now we will use the Pythagorean theorem. We know from part a that side b = 10.57, and the figure has a hypotenuse of 25:
a²+(10.57)² = 25²
a² + 111.7249 = 625
Subtract 111.7249 from both sides:
a²+111.7249-111.7249=625-111.7249
a² = 513.2751
Take the square root of both sides:
√a² = √513.2751
a = 22.66
#2)
A) Let A be the 15° angle, B be the angle to the right and C be the angle at the top of the triangle. This means side a is across from angle A, side B is across from angle B, and side c is across from angle C.
Using the law of cosines,
a²=3²+4²-2(3)(4)cos(15)
a²=9+16-24cos(15)
a²=25-24cos(15)
a²=1.8178
Take the square root of both sides:
√a² = √1.8178
a = 1.3483≈1.35
Now we can use the Law of Sines to find angle C:
sin 15/1.35 = sin C/4
Cross multiply:
4*sin 15 = 1.35* sin C
Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin C)/1.35
(4*sin 15)/1.35 = sin C
Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin C)
sin⁻¹((4*sin 15)/1.35) = C
50.07 = C
To find angle B, add angle A and angle C together and subtract from 180:
B=180-(50.07+15) = 180-65.07 = 114.93
B) To find angle A, add angle B and angle C together and subtract from 180:
180-(52+45) = 180-97 = 83
Now use the Law of Sines to find side b (across from angle B):
sin 52/12 = sin 45/b
Cross multiply:
b*sin 52 = 12*sin 45
Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
b = 10.77
Find side a using the Law of Sines:
sin 83/a = sin 52/12
Cross multiply:
12*sin 83 = a*sin 52
Divide both sides by sin 52:
(12*sin 83)/(sin 52) = (a*sin 52)/(sin 52)
15.11 = a
