Can someone please help actually and not put a random answer please it’s due at 12:00 I’ll give brainliest

1)(-3,4)

2)2

3)(x-0)^2+(y+5)^2=14

Step-by-step explanation:

1) so the center of a circle equation is (x-h)^2+(y-k)^2=r^2

where (h,k) are center of the circle

so as h we have 3but since its -h we must take -3 as x coordinate for center

and we need to take 4 as y coordinate since its -k and k=4 so-(-4)=4

center of circle: (-3,4)

since r^2represents the radius in a circle, the r^2 value is 4

to find just r we must take the square root

sqrt of 4=2

hence radius is 2

(x-0)^2+(y--5)^2=14

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.