Find the equation of a line that passes through the point (0, -2) and is parallel to a line that goes through the points (4,0) and (0,3).

3x + 4y +8 = 0

Step-by-step explanation:

(X1,Y1) = (4,0)

(X2,Y2)=(0,3)

(X3,Y3)=(0, - 2)

Slope(m)=(Y2-Y1)/(X2-X1)

= (3-0)/(0-4)

= -3/4

Equation of the line,

Y-Y3 = m(X-X3)

Y-(-2)= (-3/4) × (X-0)

Y+2= (-3/4)X

Y+2= -3X/4

4(Y+2)= -3X

4Y+8= -3X

4Y+3X+8=0 is the required equation.

definition: a circle is the  set of all points in a plane that are equidistant  from a given  point  called the center of the circle. we use the symbol ⊙ to represent a circle. the a line segment from the center of the circle to  any point  on the circle is a radius of the circle.

step-by-step explanation:

i googled it and this is what i got i don't know how it is. sorry.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2