1) What is the equation of the line that has a slope of 2/3 and
goes through (0, -4)?

2x-3y-12=0

Step-by-step explanation:

solution:

slope(m) = 2/3

(x1 , y1) = (0, -4)

we know that,

(y -y1) = m(x - x1)

or, y - (-4) = 2/3(x - 0)

or, y + 4 = 2/3 *x

or, 3(y + 4) = 2x

or, 3y + 12 = 2x

2x - 3y -12 = 0 is the required equation of line.

step-by-step explanation:

1. the cylinder is a. 54

2.   a.108 /

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