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vovikov84 [41]
3 years ago
11

Dado dos ángulos complementarios, uno mide (2x + 10) y el otro (x + 20),

Mathematics
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

El valor de <em>x</em> es igual a 20 o <em>x</em> = 20.

Step-by-step explanation:

Lo primero que se debe saber es que <em>dos ángulos complementarios suman un ángulo recto o 90º</em>.

Supongamos que el valor de un ángulo \\ \alpha y un ángulo \\ \beta valen:

\\ \alpha = 2x + 10 [1]

\\ \beta = x + 20 [2]

Como la suma de  \\ \alpha + \beta = 90 [3]

Entonces

\\ \alpha + \beta = (2x + 10) + (x + 20) = 90

Sumamos los factores comunes entre si:

\\ (2x + x) + (10 + 20) = 90

Para la primera expresión debemos recordar que se suman sólo los coeficientes. Así:

\\ (2 + 1)x + (10 + 20) = 90

\\ 3x + 30 = 90

Para despejar la incógnita <em>x</em>, debemos tener en cuenta que <em>una igualdad no se altera si se suma, se resta, se multiplica o divide un mismo valor a cada lado de ella</em>. Por esta razón, para despejar 3x, lo primero que podemos hacer es sumar -30 a cada lado de la expresión (lo que es igual a restar 30 a cada lado de la misma). Así tenemos:

\\ 3x + 30 - 30 = 90 - 30

\\ 3x + 0 = 90 - 30

\\ 3x = 60

Ahora dividimos cada miembro de la igualdad entre 3 (o multiplicamos cada lado de la igualdad por \\ \frac{1}{3} ):

\\ \frac{3}{3}x = \frac{60}{3}

Como sabemos que:

\\ \frac{3}{3} = 1

Entonces:

\\ 1*x = \frac{60}{3}

\\ x = \frac{60}{3}

\\ x = 20

De esta manera, el valor de <em>x</em> es igual a 20 o x = 20.

Lo anterior lo podemos comprobar considerando las ecuaciones [1], [2] y [3]. Así tenemos que:

\\ \alpha = 2x + 10 [1]

Sustituimos x por el valor de 20:

\\ \alpha = 2*20 + 10 = 40 + 10 = 50

\\ \beta = x + 20 [2]

Hacemos lo mismo para [2]:

\\ \beta = 20 + 20

\\ \beta = 40

De esta manera:

\\ \alpha + \beta = 90 [3]

\\ 50 + 40 = 90

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The material used to make a storage box costs $1.25 per square foot. The boxes have the same volume, so how much does a company
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Answer: A company saves $134.38 by choosing to make 50 of Box 2 instead of Box 1

Step-by-step explanation:

Hi, to answer this question we have to calculate the surface area of both boxes.

<em>Surface area (A) = 2wl + 2lh + 2hw, </em>

Where w is width, l is length , and h is height.

  • Box 1 = A = 2(6x20) + 2( 20x4) + 2 (4 x 6) =2 (120)+ 2 (80)+ 2(24) =448 in2
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Since 1 inch = 0.0833333 foot

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Since the material used to make a storage box costs $1.25 per square foot.

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3 years ago
For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[
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The following are the solution to the given points:

Step-by-step explanation:

Given value:

1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

Solve point 1 that is \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\:

when,

k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

                  = \frac{1}{2} - \frac{1}{3}\\\\

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

                  = \frac{1}{3} - \frac{1}{4}\\\\

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

when,

k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

                  = \frac{1}{8 \times 9}\\\\= \frac{1}{72}

k= 3 \to  s_1 = \frac{1}{(3+6)(3+7)}\\\\

                  = \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\

k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

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