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zubka84 [21]
3 years ago
12

The longest side of an acute isosceles triangle is 12 centimeters. Rounded to the nearest tenth, what is the smallest possible l

ength of one of the two congruent sides?
Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0

Answer:

The smallest possible length would be 8.5 cm.

Step-by-step explanation:

Since, an acute isosceles triangle having two congruent sides with three acute interior angles,

Given,

The longest side of an acute isosceles triangle is 12 centimeters,

Let x be the side length of the each of the congruent sides,

So, by the property of acute isosceles triangle,

x^2+x^2\geq (12)^2

2x^2\geq 144

x^2\geq 72

\implies x\geq \sqrt{72}=8.48528137424\approx 8.5

Hence, the smallest possible length of one of the two congruent sides is 8.5 cm

liraira [26]3 years ago
6 0
12/sqrt (2)=6/sqrt (2), 8.484cm. Round it up to 8.5cm
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Si el máximo común divisor (MCD) de 6432 y 132 disminuye en 8, entonces será igual a: * 1 punto 6 4 -2 -6
swat32

Answer:

4

Step-by-step explanation:

Para empezar el maximo común divisor (MCD) de dos números se refiere al número en común que tienen ambos números, entre los cuales pueden ser divididos. En el caso del máximo se refiere al número más alto posible en que ambos números pueden ser divididos.

Para poder saber esto, es necesario anotar los factores en los cuales pueden ser divididos ambos números. Sin embargo, ya que son cantidades altas, puede tomar mucho tiempo determinar esto, por lo tanto nos vamos a la respuesta final.

Sabemos que el MCD de 132 y 6432 es disminuido en 8 unidades, es decir, se le resta 8 al MCD de esos números, y ese número debe coincidir con alguna de las opciones puesta ahí. Por lo tanto, lo que vamos a hacer para resolverlo rápidamente es sumarle 8 a cada una de las opciones y luego, verificaremos si el número obtenido es divisor de 132 y 6432.

<u>Obtención del MCD original:</u>

a) 6 + 8 = 14

b) 4 + 8 = 12

c) 2 + 8 = 6

d) 6 + 8 = 2

Con estos resultados, veamos cual de ellos es el MCD de 132 y 6432.

<u>Para el 6432:</u>

6432 / 2 = 3216

6432 / 6 = 1072

6432 / 12 = 536

6432 / 14 = 459,24

Podemos observar con estos resultados que el 14 queda descartado al tener un resultado decimal, por lo tanto debe ser 2, 6 o 12. Veamos la cuenta con el 132:

<u>Para el 132:</u>

132 / 2 = 66

132 / 6 = 22

132 / 12 = 11

Podemos observar ahora en este caso que el 132 es divisible entre 12, por lo tanto, es el MCD entre 132 y 6432, asi que se concluye que el resultado de disminuir por 8 el MCD de ambos números es:

<h2>12 - 8 = 4</h2>

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Step-by-step explanation:

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Which inequality is shown in the graph?
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2 years ago
A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11
mixer [17]
Let A(t) denote the amount of salt in the tank at time t. We're given that the tank initially holds A(0)=100 lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}
\implies A'(t)+\dfrac{11}{200+33t}A(t)=484

Find the integrating factor:

\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}

Distribute \mu(t) along both sides of the ODE:

(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}
\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}
A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt
A(t)=22(200+33t)^{2/3}+C

Since A(0)=100, we get

100=22(200)^{2/3}+C\implies C\approx-652.39

so that the particular solution for A(t) is

A(t)=22(200+33t)^{2/3}-652.39

The tank becomes full when the volume of solution in the tank at time t is the same as the total volume of the tank:

200+(44-11)t=500\implies 33t=300\implies t\approx9.09

at which point the amount of salt in the solution would be

A(9.09)\approx733.47\text{ lb}
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3 years ago
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