Identify the solution to the system of equations 4x-y=7 -12x+3y=-21
1 answer:
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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.
Answer:
E.
Step-by-step explanation:
Consider a circle having centre P and radius 20 inches,
So, the area of the circle,
Also, suppose points Q and R are on the circumference,
i.e. PQ = PR = 20 inches,
If QR = 20 inches,
So, triangle PQR is an equilateral triangle,
⇒ m∠RPQ = 60°,
Now, suppose R' is another point on the circle,
Such that, ΔPQR ≅ Δ PQR',
⇒ m∠QPR' = 60°,
Thus, minor angle, m∠RPR' = m∠RPQ + m∠QPR' = 60° + 60° = 120°,
⇒ major angle, m∠RPR' = 360° - 120° = 240°,
So, the area of the circle where a point on the circumference is closer to P than it is to Q
Hence, the probability that a point is closer to P than it is to Q =
i.e. OPTION E is correct.
