What is the solution for x in the equation? -x + 3/7 = 2x -25/7
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Answer:
Step-by-step explanation:
You know how subtraction is the <em>opposite of addition </em>and division is the <em>opposite of multiplication</em>? A logarithm is the <em>opposite of an exponent</em>. You know how you can rewrite the equation 3 + 2 = 5 as 5 - 3 = 2, or the equation 3 × 2 = 6 as 6 ÷ 3 = 2? This is really useful when one of those numbers on the left is unknown. 3 + _ = 8 can be rewritten as 8 - 3 = _, 4 × _ = 12 can be rewritten as 12 ÷ 4 = _. We get all our knowns on one side and our unknown by itself on the other, and the rest is computation.
We know that ; as a logarithm, the <em>exponent</em> gets moved to its own side of the equation, and we write the equation like this:
, which you read as "the logarithm base 3 of 9 is 2." You could also read it as "the power you need to raise 3 to to get 9 is 2."
One historical quirk: because we use the decimal system, it's assumed that an expression like uses <em>base 10</em>, and you'd interpret it as "What power do I raise 10 to to get 1000?"
The expression means "the power you need to raise 10 to to get 100 is x," or, rearranging: "10 to the x is equal to 100," which in symbols is
.
(If we wanted to, we could also solve this: , so
)
El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.
<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>
La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.
Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:
σ = W / (π · D² / 4)
Donde:
- W - Peso de la caja, en newtons.
- D - Diámetro del área transversal de la caja, en metros.
Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:
σ = 25 N / [π · (0.02 m)² / 4]
σ ≈ 79577.472 Pa
<h3>Observación</h3>La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.
Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145
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