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Consider this function.
f(x) = lx - 4 + 6
If the domain is restricted to the portion of the graph with a positive slope, how are the domain and range of the function
and its inverse related?
O Since the domain of the original function is limited to x > 6, the range of the inverse function is y s 6.
O Since the domain of the original function is limited to x > 4, the range of the inverse function is y s 1.
O Since the range of the original function is limited to y > 6, the domain of the inverse function is x 2 6.
O Since the range of the original function is limited to y > 4, the domain of the inverse function is x 21.

69,69,69,69 because it is 69

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

omg what happend to your question? if you could edit this question i can

step-by-step explanation: