If the endpoints of a diameter are (6,3) and (2,1) the midpoint is the center of the circle so:
(x,y)=((6+2)/2, (3+1)/2)=(4,2)
Now we need to find the radius....the diameter is:
d^2=(6-2)^2+(3-1)^2
d^2=16+4
d^2=20 since d=2r, r=d/2, and r^2=d^2/4 so
r^2=5
The standard form of the circle is (x-h)^2+(y-k)^2=r^2 and we know:
(h,k)=(4,2) from earlier so:
(x-4)^2+(y-2)^2=5
Answer:
blank 1: toby
blank 2: 4
Step-by-step explanation:
if im right i dont want ur money
Answer:
y = x + 3
Step-by-step explanation:
the equation ofa line in slope-intercept form is
y = mx + c ( m is the slope and c the y-intercept )
here m = 1 and c = 3, hence
y = x + 3 ← is the equation of the line
hello
the question here is
![3\sqrt[]{7}(14-4\sqrt[]{56})](https://tex.z-dn.net/?f=3%5Csqrt%5B%5D%7B7%7D%2814-4%5Csqrt%5B%5D%7B56%7D%29)
step 1
multiply through the bracket by the coeffiecient
![\begin{gathered} 3\sqrt[]{7}(14-4\sqrt[]{56}) \\ (3\sqrt[]{7}\times14)-(3\sqrt[]{7}\times4\sqrt[]{56}) \\ (14\times3\sqrt[]{7})-3\sqrt[]{7}\times4\sqrt[]{4\times14} \\ (42\sqrt[]{7})-3\sqrt[]{7}\times8\sqrt[]{14} \\ (42\sqrt[]{7})-\lbrack(3\times8)\sqrt[]{7\times14} \\ 42\sqrt[]{7}-24\sqrt[]{98} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%203%5Csqrt%5B%5D%7B7%7D%2814-4%5Csqrt%5B%5D%7B56%7D%29%20%5C%5C%20%283%5Csqrt%5B%5D%7B7%7D%5Ctimes14%29-%283%5Csqrt%5B%5D%7B7%7D%5Ctimes4%5Csqrt%5B%5D%7B56%7D%29%20%5C%5C%20%2814%5Ctimes3%5Csqrt%5B%5D%7B7%7D%29-3%5Csqrt%5B%5D%7B7%7D%5Ctimes4%5Csqrt%5B%5D%7B4%5Ctimes14%7D%20%5C%5C%20%2842%5Csqrt%5B%5D%7B7%7D%29-3%5Csqrt%5B%5D%7B7%7D%5Ctimes8%5Csqrt%5B%5D%7B14%7D%20%5C%5C%20%2842%5Csqrt%5B%5D%7B7%7D%29-%5Clbrack%283%5Ctimes8%29%5Csqrt%5B%5D%7B7%5Ctimes14%7D%20%5C%5C%2042%5Csqrt%5B%5D%7B7%7D-24%5Csqrt%5B%5D%7B98%7D%20%5Cend%7Bgathered%7D)