The answer to the problem is as follows:
x = sin(t/2)
<span>y = cos(t/2) </span>
<span>Square both equations and add to eliminate the parameter t: </span>
<span>x^2 + y^2 = sin^2(t/2) + cos^2(t/2) = 1 </span>
<span>The final step is translating the original parameter limits into limits on x and y. Over the -Pi to +Pi range of t, x varies from -1 to +1, whereas y varies from 0 to 1. Thus we have the semicircle in quadrants I and II: y >= 0.</span>
Answer: 82°
Step-by-step explanation:
6 = 98°
180° - 98° = <em><u>82°</u></em>
Answer:
see explanation
Step-by-step explanation:
Given
2BD = 7BT , then
2(d - b ) = 7(t - b) ← distribute both sides
2d - 2b = 7t - 7b ( add 7b to both sides )
2d + 5b = 7t, thus
2![\left[\begin{array}{ccc}10.5\\10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10.5%5C%5C10%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ 5![\left[\begin{array}{ccc}0\\3\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C3%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= 7t
![\left[\begin{array}{ccc}21\\20\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C20%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ ![\left[\begin{array}{ccc}0\\15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= 7t
![\left[\begin{array}{ccc}21\\35\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C35%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= 7t , thus
t = 
= ![\left[\begin{array}{ccc}3\\5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Hence T = (3, 5 )
Answer:
2 1/3
Step-by-step explanation:
