Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

x = 3 + √6 ; x = 3 - √6 ; ;  

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

for the equation ax² + bx + c = 0

thus,

the roots are

or

or

and, x =

or

and, x =

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

or

or

and, x =

or

and, x =

or

and, x =

or

and,

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; ;

answer: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere

answer: -10

step-by-step explanation:

s(t) = 1 - 10t

s(x) = 1 - 10x

s(x + h) = 1 - 10(x + h)

            = 1 - 10x - 10h

s(x + h) - s(x) = 1 - 10x - 10h - (1 - 10x)

                      = 1 - 10x - 10h - 1 + 10x

                      = -10h

= -10