Can someone PLEASE show me how to do this! If f (x) = x^2 + 3x +2/ x +3, then f’ (x) =

c

Step-by-step explanation:

The equation =( denominater * derivative of numerator - numerator * derivative of denominator) / denominator ^2

so the qstn is (x^2 + 3x +2) / (x+3)

apply the values as the above eqtn states

ie,[ (x+3) * derivative of (x^2 +3x + 2)] - [( x^2 +3x + 2) *derivative of (x+3)] /

(x+3)^2

derivative of numerator, (x^2 +3x + 2) is 2x+3

" of denominator, (x+3) is 1

so we get

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

open the brackets

[ 2x^2 + 3x + 6x + 9 - x^2 +3x + 2 ] / (x+3)^2

subtract similar terms and we get the final answer in option c

drop him sis he’s not worth your precious time

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this