Find the value of the variable in the parallelogram. RZ = 4x + 2 SW=9x - 8 X=
1 answer:
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Answer: Ground Speed = 91 km/hr, Bearing = 189°
<u>Step-by-step explanation:</u>
Step 1: Draw a picture (see attached) to determine the angle between the given vectors. Notice that I moved the wind vector 180° <em>so the head of the wind vector would line up with the tail of the plane vector. </em>This created an angle of 34° between the plane and wind vectors. <em>Why?</em>
- the dashed line is 45°
- 79° (plane) - 45° (wind) = 34°
Step 2: Solve for the length of the resultant vector using Law of Cosines
<em>c² = a² + b² - ab cos C</em>
c² = (111)² + (25)² - (111)(25) cos 34°
c² = 12,946 - 4601
c² = 8345
c = 91
Ground speed is 91 km/hr
Step 3: Solve for the bearing of the resultant vector using Law of Sines
A = 9°
<em>Reminder that we moved the wind vector 180° to create the resultant vector so we need to add 180° to our answer.</em>
Bearing = A + 180°
= 9° + 180°
= 189°
Answer:
( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 ), ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )
y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 , y + 1 = ( x + 2 ) - √ 2
y + 1 = ( x + 2 ) + √ 2 , ( x + 2 )^2 + ( y + 1)^2 = 1
Step-by-step explanation:
Given:
- Four functions to construct a diamond:
y = −x+ √ 2, y = −x− √ 2, y = x+ √ 2, and y = x − √ 2.
Find:
a)Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once.
b)Find the intersection points between the unit circle and each of the four lines.
(c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help.
Solution:
- For first part see the attachment.
- The equation of the unit circle is given as follows:
x^2 + y^2 = 1
- To determine points of intersection we have to solve each given function of y with unit circle equation for set of points of intersection:
For: y = −x+ √ 2 , x - √ 2
And: x^2 + y^2 = 1
x^2 + (+/- * (x - √ 2))^2 = 1
x^2 + (x - √ 2)^2 = 1
2x^2 -2√ 2*x + 2 = 1
2x^2 -2√ 2*x + 1 = 0
2[ x^2 - √ 2] + 1 = 0
Complete sqr: (1 - 1/√ 2)^2 = 0
x = 1/√ 2 , x = 1/√ 2
y = -1/√ 2 + √ 2 = 1/√ 2
y = 1/√ 2 - √ 2 = - 1/√ 2
Points are: ( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 )
- Using vertical symmetry of unit circle we can also evaluate other intersection points by intuition:
x = - 1/√ 2
y = 1/√ 2 , -1/√ 2
Points are: ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )
- To determine the function for the rhombus region that would be tangential to unit circle with center at ( - 2 , - 1 ):
- To shift our unit circle from origin to ( - 2 , - 1 ) i.e two units left and 1 unit down.
- For shifts we use the following substitutions:
x = x + 2 ....... 2 units of left shift
y = y + 1 .......... 1 unit of down shift
- Now substitute the above shifting expression in all for functions we have:
y = −x+ √ 2 -----> y + 1 = - ( x + 2 ) + √ 2
y = −x− √ 2 -----> y + 1 = - ( x + 2 ) - √ 2
y = x- √ 2 -------> y + 1 = ( x + 2 ) - √ 2
y = x+ √ 2 ------> y + 1 = ( x + 2 ) + √ 2
x^2 + y^2 = 1 -----> ( x + 2 )^2 + ( y + 1)^2 = 1
- The following diamond shape graph would have the 4 functions as:
y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 , y + 1 = ( x + 2 ) - √ 2
y + 1 = ( x + 2 ) + √ 2 , ( x + 2 )^2 + ( y + 1)^2 = 1
- See attachment for the new sketch.
