Here, the given equation is,
x^2+y^2+px+6y-3=0
By copairing it with x^2+y^2+2gx+2fy+c=0, we get,
2g=p or, g=p/2
2f=6 or, f=3
c=-3
now the radius is,
r^2=g^2+f^2-c
or, 4^2=(p/2)^2 +3^2 -(-3)
or, 16= p^2/4 +9+3
or, 16-9-3=p^2/4
or p^2/4=4
or, p^2=4×4
or p^2=4^2
or p=4
therefore the rewuired value of p is 4.
D because -50x is the amount you’re subtracting from the total
Your absolute value cannot be negative.
Answer:
a = 4
Step-by-step explanation:
a^2 + b^2 = c^2
a= ?
b= 3
c= 5 (c is always the hypotenuse)
*plug in given values
a^2 + 3^2 = 5^2
a^2 + 9 = 25
-9 -9
a^2 = 16
*find the square root
sqrt(a) = sqrt(16)
a = 4
Answer:x=-14.75
Step-by-step explanation:
f(x) = -7x - 19=-19
g(x) = 0.75x + 4.25 =4.25
