Mathematics, 10.11.2020 16:40 lileljusto2829

Tenemos un rectángulo cuya área es (15n + 9nx) cm 2 y su base mide (3n) cm. Existe también un cuadrado con la misma altura del rectángulo, tal que si a su área se le disminuye (9x 2 ) cm 2, , ésta medirá 1825 cm 2 . ¿Cuánto mide el área original del cuadrado?

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Answer from: taylordalton93

El área original del cuadrado es de 1825 centímetros cuadrados.

Step-by-step explanation:

De acuerdo con el enunciado, tenemos que el área del cuadrilátero es:

$(15\cdot n +9\cdot n\cdot x)\,cm^{2} = (3\cdot n\,cm)\cdot h$ (1)

Donde es la altura del cuadrilátero, medida en centímetros.

De ahí sacamos que la altura es igual a:

$h = (15+9\cdot x)\,cm$

También se sabe que el cuadrado con la misma altura del cuadrilátero menos un área determinada es igual a lo siguiente:

$(h^{2}-9\cdot x^{2})\,cm^{2} = 1825\,cm^{2}$

$[(15+9\cdot x)^{2}-9\cdot x^{2}]\,cm^{2} = 1825\,cm^{2}$

$225+270\cdot x+81\cdot x^{2}-9\cdot x^{2} = 1825$

$72\cdot x^{2}+270\cdot x +225 = 1825$

$72\cdot x^{2}+270\cdot x -1600 = 0$ (2)

Esta es una ecuación cuadrática, el cual puede ser resuelto por la Fórmula de la Cuadrática, cuyas soluciones son, respectivamente:

$x_{1} \approx 3.198$ y $x_{2}\approx -6.948$

Sabemos que las longitudes son variables positivas, por tanto, solo se debe escoger aquellas soluciones de que cumplan tal condición. Empleamos la altura del cuadrilátero para la evaluación:

$x_{1} \approx 3.198$

$h_{1} = 15+9\cdot (3.198)$

$h_{1}\approx 43.782\,cm$

$x_{2}\approx -6.948$

$h_{2} = 15+9\cdot (-6.948)$

$h_{2}\approx -47.532\,cm$

Entonces, el valor apropiado de es aproximadamente 3.198.

Por último el área original del cuadrado es:

$A = (43.782\,cm)^{2}$

$A = 1916.864\,cm^{2}$

El área original del cuadrado es de 1825 centímetros cuadrados.

Answer from: Quest

What’s the question?
Max points i can ! it’s in math because i trust the math people more.

Answer from: Quest

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

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