First, let's calculate the horizontal and vertical components of the wind speed (W) and the airplane speed (A), knowing that south is a bearing of 270° and northeast is a bearing of 45°:
Now, let's add the components of the same direction:
To find the resultant bearing (theta), we can use the formula below:
The angle -86° is equivalent to -86 + 360 = 274°.
Therefore the correct option is b.
Answer:
x = 13
Arc RQ = 180°
Step-by-step explanation:
PEQ is an inscribed angle and the measure of an inscribed angle in a circle is equal to half of the arc it sees.
The perimeter of a circle is always 360° therefore
2 × (5x + 15) + 9x + 17 + 6x - 12 = 25x + 35 = 360°
25x = 360 - 35
25x = 325
x = 13
Since arc RQ is equal to 10x + 30 we replace x with the value we just got
10×13 + 30 = 160°
Answer:
1.72
Step-by-step explanation:
AB tangent at D, AD = OD = 4
so triangle OAD is right angle with side of 4 and 4.
area of OAD = 1/2 * 4 * 4 = 8
Angle AOD = DAO = 45 deg.
so circular sector OCD area = area of circle O * 45/360
= pi * 4 * 4 * 45/360
= 2pi
Shade area ACD = trigangle OAD - circular sector OCD
= 8 - 2pi
= 1.72
Answer:
a. 0.48
b. 0.14
c. 0.47
Step-by-step explanation:
Data provided in the question
0 - 2 0.48
3 - 5 0.26
6 - 8 0.12
9- 11 0.09
12- 14 0.05
Based on the above information
a. The probability for fewer than three phobias is
= P( x < 3)
= 0.48
b. The probability for more than eight phobias is
= P( x >8)
= 0.09 + 0.05
= 0.14
c. Probability between 3 and 11 phobias is
= P(3 < x < 11)
= 0.26 + 0.12 + 0.09
= 0.47
Let
x--------> <span>the unknown scale size
we know that
N 1---------> 2.618
N 2---------> x
N 3---------> 1
N 4---------> 0.618
to obtain the size N 4 ----> multiply the size N 3 by 0.618
so
to obtain </span> the size N 2 ----> multiply the size N 1 by 0.618
size N 2=2.618*0.618------> 1.6179
x=1.618
the answer is
the unknown scale size is 1.618
