A quadratic equation has the zeros –3 and 6. Which quadratic equation can be the given equation?

It can be A. (not B or C)

Step-by-step explanation:

It is A because x-6=0 can be simplified to x=6. Then, 2x + 6, you can divide the whole equation resulting in x+3=0, simplify this and you get x=-3. YES

It is not B because, while x+3=0 results in a zero of -3, 6x-1 can be simplified to be divided by 6. When we do this we get x-1/6=0, which is not equivalent to 6. NO

It is not C because, while x-6=0 results in a zero of 6, -3x can be simplified with the zero product property to get -3x=0 then dividing -3 by 0 giving you 0 which is not equivalent to one. NO

Step-by-step explanation:

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -