Which function is negative for the interval [–1, 1]? On a coordinate plane, a curved line with a minimum value of (negative 1, negative 2), crosses the x-axis at (negative 2.5, 0) and (0.5, 0), and the y-axis at (0, negative 1). On a coordinate plane, a curved line with a minimum value of (0, negative 2), crosses the x-axis at (negative 1.5, 0) and (1.5, 0). On a coordinate plane, a curved line with a maximum value of (0, 2) and a minimum value of (1.5, negative 2.75), crosses the x-axis at (negative 0.6, 0) and (0.6, 0). On a coordinate plane, a curved line with a maximum value of (negative 1.25, 4) and a minimum value of (1.5, negative 3.5), crosses the x-axis at (negative 1.75, 0), (0.75, 0), and (1.75, 0), and crosses the y-axis at (0, 2).

As the x-values go to zero, the function's values go to positive infinity.

Step-by-step explanation:

also answer B

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

c: lkllmn.

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