A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E. There is a 10mi/h cross wind blowing due east. What is the balloon's actual speed and direction? Round angles to the nearest degree and other values to the nearest tenth. Answers:
20.9 mi / h; N 56° E
27.1 mi / h; N 53° E
27.1 mi / h; N 37° E
20.9 mi / h; N 34° E

27.1 mi / h; N 53° E

Step-by-step explanation:

Begin by finding the component form of the two vectors.

The balloon's vector has a magnitude of 20 mi/h and makes an angle of 54° with the x-axis.

The figure shows a vector of balloon speed on a Cartesian plane. The initial point of the vector is the origin. The vector of balloon speed has a length of 20 units. The angle between the vector and the positive direction of the x axis measures 54 degrees.

The balloon's vector components can be determined from the length of the hypotenuse and the angle made with the x-axis.

cos54°=x20

Multiply both sides by 20.

20(cos54°)=x

Calculate and round to the nearest tenth.

11.7≈x

sin54°=y20

Multiply both sides by 20

20(sin54°)=y

Calculate and round to the nearest tenth.

16.2≈y

The balloon's vector is <11.7,16.2>.

Since the wind blows 10 mi/h in the direction of the positive x-axis, it has a horizontal component of 10, and a vertical component of 0.

The figure shows a vector of wind speed on a Cartesian plane. The initial point of the vector is the origin. The vector of wind speed is directed along the positive x axis and has a length of 10 units.

The wind's vector is <10,0>.

To combine the vectors, add the components of the balloon's vector and the wind's vector .

<11.7,16.2>+<10,0>=<21.7,16.2>

The resultant vector in component form is <21.7,16.2>.

The magnitude of the resultant vector is the balloon's actual speed. To find the magnitude of the resultant vector, use the Distance Formula.

∣∣<21.7,16.2>∣∣=(21.7)2+(16.2)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

Simplify.

∣∣<21.7,16.2>∣∣=470.89+262.44‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

Simplify again.

∣∣<21.7,16.2>∣∣=733.33‾‾‾‾‾‾‾√

Simplify the square root and round to the nearest tenth.

∣∣<21.7,16.2>∣∣≈27.1 mi/h

The figure shows a vector of balloon speed, a vector of wind speed and a resultant vector on a Cartesian plane. The initial point of the vector of balloon speed is the origin. The initial point of the vector of wind speed is the terminal point of the vector of balloon speed. The initial point of the resultant vector is the origin. The terminal point of the resultant vector is the terminal point of the vector of wind speed. The angle between the resultant vector and the positive direction of x axis is labeled as A.

The angle A is formed by the resultant vector and the x-axis and gives the balloon's actual direction. The tangent of the angle formed with the x-axis is the y component over the x component.

tanA=16.221.7

Solve for m∠A

m∠A=tan−1(16.221.7)

Round to the nearest degree.

m∠A≈37°

This is equivalent to the direction N 53°E, the angle formed by the resultant vector and the y-axis.

Therefore, the balloon's actual speed is 27.1 mi/h, and actual direction is N 53°E.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.