Es la opción D, y al altura del Triángulo es de 6.92cm
Respuesta:
Pregunta № 3
[tex] {x}^{2} = {12}^{2} + {35}^{2} [/tex]
[tex] {x}^{2} = 144 + 1225[/tex]
[tex] {x}^{2} = 1369[/tex]
[tex] \sqrt{ {x}^{2} } = \sqrt{1369} [/tex]
[tex]x = 37m[/tex]
Literal D ✓
Pregunta № 4 .
[tex] {x}^{2} = {8}^{2} - {4}^{2} [/tex]
[tex] {x}^{2} = 64 - 16[/tex]
[tex] {x}^{2} = 48[/tex]
[tex] \sqrt{ {x}^{2} } = \sqrt{48} [/tex]
[tex]x = 4 \sqrt{3} cm[/tex]
Literal A ™
Saludos
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Es la opción D, y al altura del Triángulo es de 6.92cm
Verified answer
Respuesta:
Solución del problema por el teorema de pitagoras.
Pregunta № 3
[tex] {x}^{2} = {12}^{2} + {35}^{2} [/tex]
[tex] {x}^{2} = 144 + 1225[/tex]
[tex] {x}^{2} = 1369[/tex]
[tex] \sqrt{ {x}^{2} } = \sqrt{1369} [/tex]
[tex]x = 37m[/tex]
Literal D ✓
Pregunta № 4 .
[tex] {x}^{2} = {8}^{2} - {4}^{2} [/tex]
[tex] {x}^{2} = 64 - 16[/tex]
[tex] {x}^{2} = 48[/tex]
[tex] \sqrt{ {x}^{2} } = \sqrt{48} [/tex]
[tex]x = 4 \sqrt{3} cm[/tex]
Literal A ™
Saludos