existe um retangulo cujo o perimetro é 40 m e cuja aerea é 44 m2 ? JUSTIFIQUE SUA RESPOSTA
2x+2y=40m
x+y=20m
xy=44m²
x=20-y
20-y ,y=44
20y-y²=44
-y²+20y-44=0
y²-20y+44=0
y= 20+-²V400-176 /2
y=20+-²V224 /2
y=20+14,966629 /2=17,483314
y=20-14,966629/2=2,5166855
p=a+a+b+b
40=2a+2b
20=a+b
a=b-20
A=(b*a)
44=b(b-20)
44=b²-20b
b²-20b-44=0
Por Bhaskara:
b'=22......?????
b''=-2...não serve
a+22=20
a=20-22
a=-2.....?????
Ola lucas
x² - 20x + 44 = 0
d² = 400 - 176 = 224 = 16*14
d = 4√14
x = (20 + 4√14)/2 = 10 + 2√14
y = 20 - 10 - 2√14 = 10 - 2√14
pronto
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2x+2y=40m
x+y=20m
xy=44m²
x=20-y
20-y ,y=44
20y-y²=44
-y²+20y-44=0
y²-20y+44=0
y= 20+-²V400-176 /2
y=20+-²V224 /2
y=20+14,966629 /2=17,483314
y=20-14,966629/2=2,5166855
p=a+a+b+b
40=2a+2b
20=a+b
a=b-20
A=(b*a)
44=b(b-20)
44=b²-20b
b²-20b-44=0
Por Bhaskara:
b'=22......?????
b''=-2...não serve
a+22=20
a=20-22
a=-2.....?????
Ola lucas
x² - 20x + 44 = 0
d² = 400 - 176 = 224 = 16*14
d = 4√14
x = (20 + 4√14)/2 = 10 + 2√14
y = 20 - 10 - 2√14 = 10 - 2√14
pronto