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EastWind [94]
3 years ago
15

A square is inscribed in a circle. If the area of the square is 9 in2, what is the ratio of the circumference of the circle to t

he perimeter of the square?
answer should be put like this:

c°/p^

Mathematics
1 answer:
aniked [119]3 years ago
3 0
Hello!

Given the formula for the area of a square is:

<span>A=<span>s2</span></span> where A is the Area and s is the length of the side of the square, we can find the length of one side of the square by substituting and solving:

<span>9<span> in2</span>=<span>s2</span></span>

<span><span>√<span>9<span> in2</span></span></span>=<span>√<span>s2</span></span></span>

<span>3 in=s</span>

<span>s=3 in</span>

Using the Pythagorean Theorem we can find the length of the squares diagonal which is also the diameter of the circle:

The Pythagorean Theorem states:

<span><span>a2</span>+<span>b=</span><span>c2</span></span> where a and b are legs of the triangle and c is the hypotenuse of the right triangle. In this case, both legs of the triangle are sides of the square so the are both the same length. Substituting and solving gives:

<span><span><span>(3i n)</span>2</span>+<span><span>(3i n)</span>2</span>=<span>c2</span></span>

<span>9<span> in2</span>+9<span> in2</span>=<span>c2</span></span>

<span>9<span> in2</span>×2=<span>c2</span></span>

<span><span>√<span>9<span> in2</span>×2</span></span>=<span>√<span>c2</span></span></span>

<span><span>√<span>9<span> in2</span></span></span><span>√2</span> in=c</span>

<span>3 in<span>√2</span>=c</span>

<span>c=3<span>√2</span> in</span>

We can now find the perimeter of the square and the circumference of the circle.

Formula for Perimeter of a square is:

<span>p=4s</span> where s is the length of a side of the square.

Substituting and calculating p gives:

<span>p=4×3 in</span>

<span>p=12 in</span>

Formula for the circumference of a circle is:

<span>c=2πr</span> where r is the radius of the circle.

Or,

<span>c=dπ</span> where d is the diameter of the circle. Remember: <span>d=2r</span>

Substituting and calculating c gives:

<span>c=3<span>√2</span>π in</span>

We can then write the ratio of the circumference to perimeter as:

<span><span><span>3<span>√2</span>π in</span><span>12 in</span></span>⇒</span>

<span><span><span>3<span>√2</span>π in</span><span>124 in</span></span>⇒</span>

<span><span><span><span>√2</span>π</span>4</span></span>

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BaLLatris [955]
You just want to simplify right?!

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= 6c^5d^6 / 15c^4d^2
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50. -12n^7p^5(n^2p^4) / 36n^6p^7
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5 0
3 years ago
Se encontro que la arista de un cubo es de 30cm, con un posible error en la medicion de 0.1. Utilice diferenciales para estimar
Ierofanga [76]

Answer:

a) El error máximo posible es 270 centímetros cúbicos. El error relativo asociado al volumen es 0.01. El error asociado al volumen es 1 por ciento.

b) El máximo error posible del área superficial es 36 centímetros cuadrados. El máximo error posible del área superficial es 36 centímetros cuadrados. El porcentaje de error del área superficial es 0.667 por ciento.

Step-by-step explanation:

Recordemos que el volumen y el área superficial de un cubo quedan representados por las respectivas fórmulas:

V = l^{3} (Ec. 1)

A_{s} = 6\cdot l^{2} (Ec. 2)

Donde:

l - Longitud de la arista, medida en centímetros.

A_{s} - Área superficial, medida en centrímetros cuadrados.

V - Volumen, medido en centímetros cúbicos.

a) El error máximo posible del volumen del cubo se estima por el siguiente diferencial:

\Delta V = \frac{\partial V}{\partial l}\cdot \Delta l (Ec. 3)

Donde:

\Delta V - Error máximo posible del volumen, medido en centímetros cúbicos.

\frac{\partial V}{\partial l} - Primera derivada parcial del volumen con respecto a la longitud de la arista, medida en centrímetros cuadrados.

\Delta l - Error máximo de medición, medido en centímetros.

La derivada parcial de la función de volumen es:

\frac{\partial V}{\partial l} = 3\cdot l^{2} (Ec. 4)

Ahora expandimos (Ec. 3):

\Delta V = 3\cdot l^{2}\cdot \Delta l

Si conocemos que l = 30\,cm y \Delta l = 0.1\,cm, el máximo error posible del volumen es:

\Delta V = 3\cdot (30\,cm)^{2}\cdot (0.1\,cm)

\Delta V = 270\,cm^{3}

El error máximo posible del volumen es 270 centímetros cúbicos.

Obtenemos el error relativo al dividir el resultado anterior por el volumen, es decir:

\epsilon_{V} = \frac{\Delta V}{V} (Ec. 5)

El volumen del cubo es: (l = 30\,cm)

V = (30\,cm)^{3}

V = 27000\,cm^{3}

Ahora sustituimos (Ec. 5):

\epsilon_{V} = \frac{270\,cm^{3}}{27000\,cm^{3}}

\epsilon_{V} = 0.01

El error relativo asociado al volumen es 0.01.

Por último, encontramos el porcentaje de error asociado al volumen:

\%\epsilon_{V} = 0.01\times 100\,\%

\%\epsilon_{V} = 1\,\%

El error asociado al volumen es 1 por ciento.

b) El error máximo posible del área superficial del cubo se estima por el siguiente diferencial:

\Delta A_{s} = \frac{\partial A_{s}}{\partial l}\cdot \Delta l (Ec. 6)

Donde:

\Delta A_{s} - Error máximo posible del área superficial, medido en centímetros cuadrados.

\frac{\partial A_{s}}{\partial l} - Primera derivada parcial del área superficial con respecto a la longitud de la arista, medida en centrímetros.

\Delta l - Error máximo de medición, medido en centímetros.

La derivada parcial de la función de área superficial es:

\frac{\partial A_{s}}{\partial l} = 12\cdot l (Ec. 7)

Ahora expandimos (Ec. 6):

\Delta A_{s} = 12\cdot l\cdot \Delta l

Si conocemos que l = 30\,cm y \Delta l = 0.1\,cm, el máximo error posible del área superficial es:

\Delta A_{S} = 12\cdot (30\,cm)\cdot (0.1\,cm)

\Delta A_{S} = 36\,cm^{2}

El máximo error posible del área superficial es 36 centímetros cuadrados.

Obtenemos el error relativo al dividir el resultado anterior por el volumen, es decir:

\epsilon_{A_{S}} = \frac{\Delta A_{S}}{A_{S}} (Ec. 8)

El área superficial del cubo es: (l = 30\,cm)

A_{S} = 6\cdot (30\,cm)^{2}

A_{S} = 5400\,cm^{2}

Ahora sustituimos (Ec. 8):

\epsilon_{A_{S}} = \frac{36\,cm^{2}}{5400\,cm^{2}}

\epsilon_{A_{S}} = 6.667\times 10^{-3}

El error relativo del área superficial es 6.667 × 10⁻³.

Por último, encontramos el porcentaje de error asociado al área superficial:

\%\epsilon_{A_{S}} = 6.667\times 10^{-3}\times 100\,\%

\%\epsilon_{A_{S}} = 0.667\,\%

El porcentaje de error del área superficial es 0.667 por ciento.

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3 years ago
One hundred nautical miles equals about 185 kilometers. To the nearest kilometer, how far in kilometers is 140 nautical miles?
Mademuasel [1]

Answer:

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Step-by-step explanation:

Butterfly method 100/185=140/???

185*140=25900

divide by 100

=295

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Step-by-step explanation:

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Use the number line to determine which of the following statements is true. Each tick is 1 unit. The quotient of P and S is posi
Mnenie [13.5K]

Answer:

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The absolute value is the same as long as they have the same number whether its - or +

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3 years ago
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