Let f(x)=2x^4+x^3+x^2-3x+r. For what value of r is f(2)=0?

the answer is 0

Step-by-step explanation:

r = -38

Step-by-step explanation:

For what value of r in the formula:  f(x)  =  2x4 + x3 + x2- 3x + r  does  f(2)  =  0?

Replace x with 2:  2·24 + 23 + 22- 3·2 + r  =  32 + 8 + 4 - 6 + r  =  38 + r

This value is to equal 0:   38 + r  =  0

                                                r  =  -38

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

ab will be equal to