Com 48 palitos de mesmo tamanho eu montei 13 figuras: uns triângulos e alguns quadrados. Quantos quadrados eu montei ?
ps: em forma de equação
[...]
Se for possível me expliquem pfv.
Respostas
respondido por:
15
Bom dia Lauris!
Solução!
Lado do triangulo=3
Lado do quadrado=4
![4x+3y=48\\\\\
x+y=13\\\\\\\\\\
4x+3y=48\\\\\
x+y=13.(-4)\\\\\\\\\\
4x+3y=48\\\\\
-4x-4y=-52\\\\\\\\\\
-y=-4\\\\\\
y= \dfrac{-4}{-1}\\\\\\
\boxed{y=4}](https://tex.z-dn.net/?f=4x%2B3y%3D48%5C%5C%5C%5C%5C%0Ax%2By%3D13%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A4x%2B3y%3D48%5C%5C%5C%5C%5C%0Ax%2By%3D13.%28-4%29%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A4x%2B3y%3D48%5C%5C%5C%5C%5C%0A-4x-4y%3D-52%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A-y%3D-4%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B-4%7D%7B-1%7D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7By%3D4%7D+%0A "4x+3y=48\\\\\
x+y=13\\\\\\\\\\
4x+3y=48\\\\\
x+y=13.(-4)\\\\\\\\\\
4x+3y=48\\\\\
-4x-4y=-52\\\\\\\\\\
-y=-4\\\\\\
y= \dfrac{-4}{-1}\\\\\\
\boxed{y=4}")
![x+y=13\\\\\
x+4=13\\\\\\
x=13-4\\\\\
\boxed{x=9}](https://tex.z-dn.net/?f=x%2By%3D13%5C%5C%5C%5C%5C%0Ax%2B4%3D13%5C%5C%5C%5C%5C%5C%0Ax%3D13-4%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D9%7D "x+y=13\\\\\
x+4=13\\\\\\
x=13-4\\\\\
\boxed{x=9}")
![Resposta:\\\\\\
9~~quadrados\\\\\
4~~triangulos](https://tex.z-dn.net/?f=Resposta%3A%5C%5C%5C%5C%5C%5C%0A9%7E%7Equadrados%5C%5C%5C%5C%5C%0A4%7E%7Etriangulos "Resposta:\\\\\\
9~~quadrados\\\\\
4~~triangulos")
Bom dia!
Bons estudos!
Solução!
Lado do triangulo=3
Lado do quadrado=4
Bom dia!
Bons estudos!
lauris2:
bom dia, mas são 48 palitos do mesmo tamanho e com esses 48 palitos formam-se 13 figuras. porem eu não consigo resolver a equação que se forma.
É verdade! contei 13 palitos.
respondido por:
0
Resposta:X+y=13 y=13-x
4x+3y=48
4x+3(13-x)=48
4x+39-3x=48
X=9
Y=13-9
Y=4
9 Quadrados e 4 Triangulos
Explicação passo-a-passo:
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