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lesya [120]
3 years ago
8

Si a los lados de un cuadrado se le aumenta en 3 y 5 m respectivamente la superficie del nuevo rectangulo es 440m .Encontrar el

lado del cuadrado inicial
Mathematics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

El lado del cuadrado inicial es 17 m.

Step-by-step explanation:

El área de un rectángulo se calcula a partir de los dos lados diferentes. Siendo a y b el valor de ambos lados, el área de la figura es el producto de los dos lados contiguos del rectángulo.

A los lados de un cuadrado se le aumenta en 3 y 5 m respectivamente la superficie del nuevo rectángulo es 440 m². Siendo x el lado del cuadrado inicial, entonces el área del rectángulo se puede expresar como:

(x+3)*(x+5)= 440 m²

Resolviendo:

x*x + 5*x + 3*X + 3*5= 440

x² + 8*x + 15=440

x² + 8*x + 15 -440= 0

x² + 8*x -425= 0

Una función cuadrática de la forma a*x² + b*x + c= 0 se puede resolver mediante la expresión:

\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}

En este caso, a=1, b=8 y c=425. Reemplazando:

\frac{-8+-\sqrt{8^{2}-4*1*(-425) } }{2*1}

Resolviendo:

\frac{-8+\sqrt{8^{2}-4*1*(-425) } }{2*1}= \frac{-8+\sqrt{64+1700} }{2*1}= \frac{-8+\sqrt{1764 } }{2}= \frac{-8+42 }{2}= \frac{34}{2} = 17

y

\frac{-8-\sqrt{8^{2}-4*1*(-425) } }{2*1}= \frac{-8-\sqrt{64+1700} }{2*1}= \frac{-8-\sqrt{1764 } }{2}= \frac{-8-42 }{2}= \frac{-50}{2} = -25

Como las longitudes no pueden ser negativas, <u><em>el lado del cuadrado inicial es 17 m.</em></u>

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Step-by-step explanation:

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If Angle 1 and Angle 2 are complementary angles and if the measure of Angle 1 is 20° more than the measure of Angle 2, determine
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An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What
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Answer:

Part A:

The probability that all of the balls selected are white:

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

P(D_i)=\frac{1}{6} for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}

P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}

P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}

P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}

P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}

P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0

Part A:

The probability that all of the balls selected are white:

P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find P(D_3|A)

The data required is calculated above:

P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

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Answer:

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Step-by-step explanation:

the area of the triangles:

12*10 = 120

120/2 = 60

since there are 2 triangles of the same dimensions that make up the surface area 60 + 60 = 120

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13*14 = 182

since there are 2 rectangles of the same dimensions 182 + 182 = 364

The last rectangular:

10*14 = 140

add all areas:

120 + 364 + 140 = 624

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