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PSYCHO15rus [73]
3 years ago
10

Find the value of z and the length of side RS.

Mathematics
1 answer:
Stells [14]3 years ago
6 0

<em>answer \\ z \: is \: equal \: to \: 12 \: units \\ side \: rs \: measures \: 9 \: units \: long \\ \: please \: see \: the \: attached \: picture \: for \: full \: solution \\ hope \: it \: helps</em>

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Can anybody help me please? i dont get it :(
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Answer:

ay bro dont ask me

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Step-by-step explanation:

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Choose the correct simplification of (2xy7)2(y4)3.
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Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t
Gnom [1K]

Answer:

Second class have higher marks and greater spread.

Step-by-step explanation:

First box plot  represents class first. From the first box plot, we get

\text{Minimum value }= 53,Q_1=62,Median=80,Q_3=86,\text{Maximum value }= 89

Range=Maximum-Minimum=89-53=36

IQR=Q_3-Q-1=86-62=24

Second box plot  represents class second. From the second box plot, we get

\text{Minimum value }= 56,Q_1=62,Median=74,Q_3=89,\text{Maximum value }= 96

Range=Maximum-Minimum=96-56=40

IQR=Q_3-Q-1=89-62=27

First class has greater minimum value, first quartile of both classes are same, second class has greater median, first class has greater third quartile and first class has greater maximum value. It means second class have higher marks but class first have less variation.

Second class has greater range and greater inter quartile range. It means data of second class has greater spread.

Therefore, second class have higher marks and greater spread.

3 0
3 years ago
usa el teorema de la altura para proponer como se podria construir un segmento cuya longitud sea media proporcional entre dos se
Mrac [35]
Usando el teorema de altura 

El teorema de altura relaciona la altura (h) de un triángulo rectángulo (ver figura) y los catetos de dos triángulos que son semejantes al anterior ABC, al trazar la altura (h) sobre la hipotenusa. De manera que e<span>n todo </span>triángulo rectángulo, la altura (h<span>) relativa a la </span>hipotenusa<span> es la </span>media geométrica<span> de las dos proyecciones de los </span>catetos<span> sobre la </span>hipotenusa<span> (</span>n<span> y </span>m<span>). Es decir, se cumple que:

</span>\frac{n}{h}= \frac{h}{m}

Dado que el problema establece <span>construir un segmento cuya longitud sea media proporcional entre dos segmentos de 4 y 9 cm, entonces, digamos que n = 4cm y m = 9cm tenmos que:

</span>\frac{4}{h}= \frac{h}{9}

De donde:

h= \sqrt{(4)(9)}=\sqrt{36}=6cm

¿Cómo se podria construir si los segmentos son de a cm y b cm?

Si los segmentos son de a y b cm entonces a y b son parámetros que pueden tomar cualquier valor positivo siempre que se cumpla que:

h= \sqrt{ab}

6 0
3 years ago
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