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kirill [66]
3 years ago
15

Geometry help help help.

Mathematics
2 answers:
prohojiy [21]3 years ago
8 0
1 = 2 = 3 = 90
4 = 113
5 = 6 = 180 - 113 = 67
8 = 180 - (90+67) = 180 - 157 = 23
7 = 9 = 180 -23 = 157
saveliy_v [14]3 years ago
3 0
1 = 2 = 3 = 90
4 = 113
5 = 6 = 67
8 = 23
7 = 9 = 157
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Una persona asegura que su casa tiene forma rectangular y que el perímetro de la misma es de 18 m y que además, su área es de 21
leva [86]

Considerando las fórmulas para el perímetro y el área de un rectángulo, hay que se chega en una <u>eccuación cuadrática sin solución</u>, o sea, las medidas no son posibles y la persona estaba mintiendo.

<h3>¿Cuál es la fórmula para el perímetro y el área de un rectángulo?</h3>

Considerando que las dimensiones son l y w, hay que:

  • El perímetro es: P = 2(l + w).
  • El área es: A = lw.

El <u>perímetro es de 18 m</u>, o sea:

2(l + w) = 18

l + w = 9

l = 9- w.

El <u>área es de 21 m²</u>, o sea:

lw = 21

(9- w)w = 21

-w² + 9 - 21 = 0

w² - 9w + 21 = 0

El discriminante es dado por:

D = 9² - 4 x 1 x 21 = -3.

El discriminante negativo implica que la <u>eccuación cuadrática no tiene solución</u>, o sea, las medidas no son posibles y la persona estaba mintiendo.

Puede-se aprender más a cerca de el perímetro y el área de un rectángulo en brainly.com/question/26475963

#SPJ1

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2 years ago
Five-hundred sixty two divided by seven
ladessa [460]
Five hundred and sixty two divided by seven would be,
=80.28
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Art [367]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166243

——————————

Solve the trigonometric equation:

     \mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}


Make a substitution:

     \mathsf{cos\,x=t\qquad (-1\le t\le 1)}

and the equation becomes

     \mathsf{2t^2-t-1=0}


Rewrite conveniently  – t  as  + t – 2t,  and then factor the left-hand side by grouping:

      \mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}


Factor out  2t + 1:

     \mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}


Substitute back for  t = cos x:

     \begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}


Therefore,

     \begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}

where  k  is an integer.


Solution set:   

\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}


I hope this helps. =)

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Do the ratios 2:4 and 5:10 form a proportion?<br> yes or no
Kisachek [45]

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Yes, the ratios do form a proportion.

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