3 Sarah left home at 10:00 and cycled north in
1 answer:
You might be interested in
Remember that the equation of a circle is:
Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.
Complete the square. We must do this for x² - 6x and y² - 10y separately.
x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9
Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.
Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.
y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.
Factor y² - 10y + 25 to get (y - 5)²
Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)
Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.
Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.
Complete the square. We must do this for x² - 6x and y² - 10y separately.
x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9
Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.
Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.
y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.
Factor y² - 10y + 25 to get (y - 5)²
Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)
Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.
Hi there!
v = ±
- - - - - - - - -
K = 1/2mv²
Isolate for the variable "v". We can begin by dividing both sides by 1/2. (Multiply by the reciprocal, or 2):
2 · K = 2 · (1/2mv²)
2K = mv²
Continue isolating by dividing both sides by "m":
2K / m = v²
Take the square root of both sides. Remember that the solution can either be positive or negative since there are positive and negative roots.
v = ±