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
8 players were on the team last year
Ok what's your quit-ionbro I will help
Answer:
x = 7
Length = 4 inches , Width = 15 inches
Step-by-step explanation:
<em>PART</em><em> </em><em>1</em><em> </em><em>:</em>
Area of rectangle = Length × Width
A = L × W
L = x - 3
W = x + 8
Therefore:
A = L × W
60 = ( x - 3 ) ( x + 8 )
( x - 3 ) ( x + 8 ) - 60 = 0
x^2 + 8x - 3x - 24 - 60 = 0
x^2 + 5x - 84 = 0
x^2 + 12x - 7x - 84 = 0
x ( x + 12 ) - 7 ( x + 12 ) = 0
( x + 12 ) ( x - 7 ) = 0
THEREFORE:
x = - 12 , x = 7
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<em>PART</em><em> </em><em>2</em><em> </em><em>:</em>
L = x - 3
L = ( - 12 ) - 3
L = - 15
OR
L = ( 7 ) - 3
L = 4
THEREFORE:
Length can not be a negative number. Hence, L can not equal to - 15.
L = 4 inches
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W = x + 8
W = ( 7 ) + 8
THEREFORE:
W = 15 inches
