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f(x)=2x+3
1. aria triunghiului determinat de grafic si axele de coordonate
2. S= f(1) + f(2) + ..... f(10)​

Răspuns :

Andyilye

Răspuns:

f(x) = 2x + 3

Intersecția cu axa Ox: y = 0 ⇒ 2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = -3/2 ⇒ A(-3/2;0)

Intersecția cu axa Oy: x = 0 ⇒ y = 2 · 0 + 3 = 3 ⇒ B(0; 3)

[tex]OA = \bigg|-\dfrac{3}{2}\bigg| = \dfrac{3}{2}[/tex]

[tex]OB = |3| = 3[/tex]

[tex]A_{\Delat AOB} = \dfrac{OA \cdot OB}{2} = \dfrac{\dfrac{3}{2} \cdot 3}{2} = \bf \dfrac{9}{4} \ u^2[/tex]

******

S= f(1) + f(2) + ..... f(10) = 2·1 + 3 + 2·2 + 3 + ... + 2·10 + 3 = 2·(1 + 2 + ... + 10) + 3·10 = 2·10·11:2 + 30 = 110 + 30 = 140

✍ Suma Gauss

[tex]\boxed {\boldsymbol{1 + 2 + 3 + ... + n = \dfrac{n \cdot (n + 1)}{2}}}[/tex]

Targoviste44

[tex]\bf 1.\\ \\ \it G_f\cap Oy =A(0,\ y) \Rightarrow y=f(0)=3 \Rightarrow A(0,\ 3)\\ \\ G_f\cap Ox =B(x,\ 0) \Rightarrow f(x)=0 \Rightarrow 2x+3=0\Rightarrow x=-1,5 \Rightarrow B(-1,5;\ 0)\\ \\ \\ \mathcal{A}_{AOB}=\dfrac{OA\cdot OB}{2}=\dfrac{3\cdot1,5}{2}=2,25\ \ (cm^2)[/tex]

2.

[tex]\it S=(2+3)+(4+3)+(6+3)+\ ...\ +(20+3)=(2+4+6+\ ...\ +20)+\\ \\ +3\cdot10=2(1+2+3+\ ...\ +10)+30=\not2\dfrac{10\cdot11}{\not2}=30+110=140[/tex]

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