Find the like terms.
14t + 14
Answer:
A rotation about point A
Step-by-step explanation:
Well, I suppose you meant a <em>Triangle AQR is rotated up and to the right to form Triangle AKP. (</em>Or... AKP to AQR...makes more sense)
In this Geometric Transformation we must choose, Rotation. You must choose an angle and the direction Clockwise/Anticlockwise.
In this case, the best choice would be a rotation, since a reflection would make both triangles sharing not only a point, but also a line segment.
If it was rotated about a point then this triangle would share a point R, not A.
$3.70d + $5 = $42
So $3.70 per d (d in miles) plus $5 flat rate = $42
In case you need it solved also. Distance is 10 miles.
This isn't an identity, so I assume you have to solve the equation.
(1 - sin(2<em>A</em>)) (1 + cot(2<em>A</em>)) = cot(2<em>A</em>)
1 - sin(2<em>A</em>) + cot(2<em>A</em>) - sin(2<em>A</em>) cot(2<em>A</em>) = cot(2<em>A</em>)
1 - sin(2<em>A</em>) - cos(2<em>A</em>) = 0
sin(2<em>A</em>) + cos(2<em>A</em>) = 1
Multiply both sides by 1/√2, which we want to do because cos(<em>π</em>/4) = sin(<em>π</em>/4) = 1/√2. This gives
cos(<em>π</em>/4) sin(2<em>A</em>) + sin(<em>π</em>/4) cos(2<em>A</em>) = 1/√2
Then condense the left side as
sin(2<em>A</em> + <em>π</em>/4) = 1/√2
2<em>A</em> + <em>π</em>/4 = sin⁻¹(1/√2) + 2<em>nπ</em> <u>or</u> 2<em>A</em> + <em>π</em>/4 = <em>π</em> - sin⁻¹(1/√2) + 2<em>nπ</em>
(where <em>n</em> is any integer)
2<em>A</em> + <em>π</em>/4 = <em>π</em>/4 + 2<em>nπ</em> <u>or</u> 2<em>A</em> + <em>π</em>/4 = 3<em>π</em>/4 + 2<em>nπ</em>
2<em>A</em> = 2<em>nπ</em> <u>or</u> 2<em>A</em> = <em>π</em>/2 + 2<em>nπ</em>
<em>A</em> = <em>nπ</em> <u>or</u> <em>A</em> = <em>π</em>/4 + <em>nπ</em>
