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3 years ago

() + -8= -6 what is the awnser

Mathematics

1 answer:

Mariana [72]3 years ago

5 0

Answer:x=2

Step-by-step explanation:

change () to x

x + -8= -6

-6+8=2

x=2

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A girl's feet are negative 2 over 4 yards from the surface of a pool. A boy's feet are negative 3 over 4 yards from the surface

frozen [14]

The girls are closer. Her feet are 1/2 the distance away from the surface pool, the boys are 3/4 away.

6 0

2 years ago

Si un rectángulo con perímetro de 74 pulgadas tiene un largo de 5 pies más que el ancho,

baherus [9]

Perímetro del rectángulo = 74 pulgadas

Deje que el ancho del rectángulo sea 'x'.

Entonces, la longitud de este rectángulo = x + 5

We know that :

$\color{hotpink}\tt \: Perímetro \: del \: rectángulo =\color{plum} 2 × (l + b)$

Lo que significa :

$=\tt 2(x + x + 5) = 74$

$=\tt 2(2x + 5) = 74$

$=\tt 4x +10 = 74$

$=\tt 4x = 74 - 10$

$=\tt 4x = 64$

$=\tt x = \frac{64}{4}$

$\color{plum} =\tt x = 16$

Por lo tanto, el ancho de este rectángulo = 16 pulgadas

Entonces, la longitud de este rectángulo :

$=\tt x + 5$

$= \tt16 + 5$

$\color{plum} =\tt 21 \: pulgadas$

La longitud de este rectángulo = 21 pulgadas

<h2>Por lo tanto :</h2>

● Longitud del rectángulo = <em>21 pulgadas</em>

● Ancho del rectángulo = <em>16 pulgadas</em>

3 0

2 years ago

What the recursive formula for the sequence

klemol [59]

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})$

so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

$\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}$

just a quick note on notation:

$\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}$