The length of a rectangle is 13 centimeters less than five times its width. its area is 28 square centimeters. find the dimensions of the rectangle. the width is nothing ▼ and the length is nothing ▼

we have given that length of a rectangle is 13 cm less that five times its width.

where l is length and b is width.

we need to set up an equation according to the problem statement.

l=5b-13

we have area =28cm^2

we know that l*b=28

(5b-13)*b=28

5b^2-13b-28=0

we need to solve it by formula

we get (5b-20)(5b+7)=0

b=4 and b=-7/5 so negative value cannot be chosen so b=4

and l=5b-13=20-13=7.

l=7

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

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: