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kumpel [21]
3 years ago
9

a point moves around the circle x^2+y^2=9. When the point is at (-sqrt3, sqrt6), the x coordinate is increasing at the rate of 2

0 units per second. How fast is its y coordinate at that instant?
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
4 0
To answer, subtract x² from both sides of the equation to give us,
                                   y² = -x² + 9
Then, we derive the equation in terms t.
                             2y(dy/dt) = -2x(dx/dt) 
Substituting the known values from the given,
                           2(sqrt6)(dy/dt) = -2(-sqrt3)(20)
The value of dy/dt is 14.14. 
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dolphi86 [110]

Answer:

With a center of (2, -2) the dilation coefficient is 3

Lt me know if I can explain any part more to help you understand.  

Step-by-step explanation:

To get the coefficient we want to see where the points start and where they end up.  it is difficult to tell the exact point of E and G and their transformations, so we will only use H and F and their transformations.  We do need 2 to get the center.

From now on I will only be oooking at y values since we know the center x value.

We know the center has to be between 0 and - 3 because it has to be inside of both shapes.

we also know that the distance from the center point to H is c - 0 and F is c - (-3).

then we also know the same number is multiplying these distances to get the new coordinates, 4 and -5 rspectively.  or more accurtely, the distance is being multiplied by the same number.

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Gala2k [10]

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