Answer:
-3.74
Step-by-step explanation:
Answer:
1/2
Step-by-step explanation:
The "Pythagorean relation" between trig functions can be used to find the sine.
<h3>Pythagorean relation</h3>
The relation between sine and cosine is the identity ...
sin(x)² +cos(x)² = 1
This can be solved for sin(x) in terms of cos(x):
sin(x) = √(1 -cos(x)²)
<h3>Application</h3>
For the present case, using the given cosine value, we find ...
sin(x) = √(1 -(√3/2)²) = √(1 -3/4) = √(1/4)
sin(x) = 1/2
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<em>Additional comment</em>
The sine and cosine of an angle are the y and x coordinates (respectively) of the corresponding point on the unit circle. The right triangle with these legs will satisfy the Pythagorean theorem with ...
sin(x)² + cos(x)² = 1 . . . . . . where 1 is the hypotenuse (radius of unit circle)
A calculator can always be used to verify the result.
Answer:
Line of east wedge is: 2x - y = 96
So, Option 1 is correct.
Step-by-step explanation:
The east edge cannot intersect with the west edge means that two lines are parallel.
If the two lines are parallel then they have same slope. We need to find the slopes of given lines and check which line has slope same as slope of west edge.
Slope of west edge.
y = 2x + 5
The standard equation for slope intercept form is:
y = mx+b
where m is the slope. So, m= 2
Now finding line for east edge.
Option 1.
Convert each given equation to standard slope intercept form and find the slope.
2x -y =96
-y = -2x +96
Multiply with -1
y = 2x -96
m = 2
Option 2.
-2x -y = 96
-y = 2x +96
y = -2x-96
m = -2
Option 3
-y-2x =48
-y = 2x +48
y = -2x -48
m = -2
Option 4.
y+2x = 48
y = -2x+48
m = -2
So, only line of Option 1 has slope = 2 which is equal to the slope of west edge.
Line of east wedge is: 2x - y = 96
So, Option 1 is correct.
Answer:
103
Step-by-step explanation:
it has to add up to 180 so
180-77=103
Answer:
∠2 and ∠5
Step-by-step explanation:
we know that
<u>Alternate Exterior Angles</u> are a pair of angles on the outer side of each of those two lines but on opposite sides of the transversal
In this problem
∠12 and ∠2 are alternate exterior angles
∠12 and ∠5 are alternate exterior angles
therefore
∠2 and ∠5 are each separately alternate exterior angles with ∠12
