PLEASE HELP QUICK WILL MARK BRAINLIEST 4) There is a square and a rectangle. The length of the rectangle is two centimeters more
than the sides of the square. The width of the rectangle is one centimeter less than the
square's sides. The sum of the areas of the two figures is thirty-four square centimeters,
Find the dimensions of each figure.

given/equations formed from text:

I: l=w+3

II: A=l*w

III: new area B=(l-2)*w

IV: B=A-30

substitute l in II with I to remove l:

II': A=l*w=

(w+3)*w=

w^2+3w

substitute l in III with I to remove l:

III': B=(l-2)*w=

(w+3-2)*w=

(w+1)*w=

w^2+w

substitute A and B from II' and III' into IV:

B=A-30

w^2+w=w^2+3w-30

30=2w

w=15

insert w=15 into I:

l=15+3=18

->

A) dimensions of full size rectangle are width 15 and length 18

dimensions of reduced size rectangle are width 15 and length 16

B) full size: 15*18=10*18+5*18=180+90=270cm^2

reduced size: 15*16=10*16+5*16=160+80=240cm^2

THANKS

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Step-by-step explanation:

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: