1 is 110. 2 is 65 and 3 is 35. You have to double 55 for the first question. For 2 you divide 2 from 130. And the third one shows the angle.
The answer would have to be d
Answer:
a) The mass is released at t = 0 when h is minimum. Half a cycle later h reaches its maximum and another half a cycle it reaches its minimum again. Hence over one cycle, h varies with t as follows:
b) According to the graph obtained in part a), h(t) could be modeled by a cosine function shifted (translated) vertically up and horizontally to the right. Hence
Step-by-step explanation:
1 and 24
2 and 12
3 and 8
4 and 6
Answer:
Equation = (x - 6 )² + ( y + 3 )² = 9
Step-by-step explanation:
The circle passes through ( 6, 0) and ( 6 , -6)
They are the coordinates of the diameter.
Using this we can find the centre of the circle.
<u>Find the centre of the circle.</u>
Centre of the circle is the mid- point of (6, 0) and ( 6, -6)
![Centre = (\frac{x_1+x_2}{2} , \frac{y_1 + y_2}{2})](https://tex.z-dn.net/?f=Centre%20%3D%20%28%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%20%2C%20%5Cfrac%7By_1%20%2B%20y_2%7D%7B2%7D%29)
![=(\frac{6 + 6}{2}, \frac{0 + (-6)}{2})\\\\=(6, -3)](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B6%20%2B%206%7D%7B2%7D%2C%20%5Cfrac%7B0%20%2B%20%28-6%29%7D%7B2%7D%29%5C%5C%5C%5C%3D%286%2C%20-3%29)
<u>Find the radius of the circle.</u>
![Radius = \frac{Diameter }{2}](https://tex.z-dn.net/?f=Radius%20%3D%20%5Cfrac%7BDiameter%20%7D%7B2%7D)
Diameter is the distance between the points (6 , 0) and ( 6, - 6)
![Diameter = \sqrt{(x_2 - x_1)^2+ (y_2 - y_1)^2\\}](https://tex.z-dn.net/?f=Diameter%20%3D%20%5Csqrt%7B%28x_2%20-%20x_1%29%5E2%2B%20%28y_2%20-%20y_1%29%5E2%5C%5C%7D)
![=\sqrt{(6-6)^2 + (-6 -0)^2}\\\\=\sqrt{0 + 36} \\\\= 6](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%286-6%29%5E2%20%2B%20%28-6%20-0%29%5E2%7D%5C%5C%5C%5C%3D%5Csqrt%7B0%20%2B%2036%7D%20%5C%5C%5C%5C%3D%206)
Therefore,
![Radius ,r = \frac{6}{2} = 3](https://tex.z-dn.net/?f=Radius%20%2Cr%20%3D%20%5Cfrac%7B6%7D%7B2%7D%20%3D%203)
<u>Standard equation of a circle:</u>
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Therefore , equation of the circle ;
![(x - 6)^2 + (y + 3)^2 = 3^2\\\\(x -6)^2 + (y + 3)^2 = 9](https://tex.z-dn.net/?f=%28x%20-%206%29%5E2%20%2B%20%28y%20%2B%203%29%5E2%20%3D%203%5E2%5C%5C%5C%5C%28x%20-6%29%5E2%20%2B%20%28y%20%2B%203%29%5E2%20%3D%209)