Notifiche
Cancella tutti

[Risolto] matemática es 11

  

0
image

 ho bisogno come si calcolano le aree degli ultimi due grazie 

Autore
7 Risposte



2
image

area parte colorata :

A = (3x+2)*(6x+3)-9x-3

A = 18x^2+12x+3 = 3(6x^2+4x+1) 

@remanzini_rinaldo 

👍 👍 👍 



3
image

Foto dritta!!!

2·(2·(2·x + 3 + 2)) - 2^2 = 8·x + 16

(3·x + 2)^2 + (3·x + 1)·(3·x + 2 - 3) = 18·x^2 + 12·x + 3

(3·x + 5)·(2·x + 3) - (3·x + 5 - 4)·(2·x + 3 - 2) = 14·x + 14

@lucianop 👍👌👍



3
image

Area colorata A = (2x+5)^2-(2x+3)^2 = 

A = 25+20x-9-12x = 8x+16 = 8(2+x) 

@remanzini_rinaldo 

👍 👍 👍 



2
11.2

=============================================================

11/2

Area parte colorata:

$\small A= (3x+2)(3x+2)+(3x+1)(3x+2-3)$

$\small A= 9x^2+6x+6x+4+(3x+1)(3x-1)$

$\small A= 9x^2+12x+4+9x^2\cancel{-3x}\cancel{+3x}-1$

$\small A= 18x^2+12x+3$

@gramor 👍👌👍



2
image

area parte colorata 

(2x+3)(3x+5)-(3x+1)(2x+1) = 14x+14 = 14(x+1)

@remanzini_rinaldo 

👍 👍 👍 



2
image

(a+b) = 75/2 m

a*b = 250 m^2

(250/b+b) = 75/2

(250+b^2) = 75b/2

250+b^2-75/2b = 0

b = (75/2±√75^2/4-1000)/4 = 8,672 ; a = 28,828 m

 

A' = (a+4)^2*(b+4)^2 = 12,672*32,828 = 416,00 m^2



1
11.3

============================================================

11/3

Area parte colorata:

$\small A= (2x+3)(3x+5)-(3x+5-2·2)(2x+3-2·1)$

$\small A= 6x^2+10x+9x+15-(3x+5-4)(2x+3-2)$

$\small A= 6x^2+19x+15-(3x+1)(2x+1)$

$\small A= 6x^2+19x+15-(6x^2+3x+2x+1)$

$\small A= 6x^2+19x+15-(6x^2+5x+1)$

$\small A= \cancel{6x^2}+19x+15\cancel{-6x^2}-5x-1$

$\small A= 19x+15-5x-1$

$\small A= 14x+14$

@gramor 👍👌👍

@remanzini_rinaldo - Grazie Rinaldo, cordiali saluti.



Risposta
SOS Matematica

4.6
SCARICA