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Justify that (x-3)(69x-2) = 69x² - 209x + 6. Determine the values of the variable x for which P exists. For x != -3 and x != 3, show that P(x+3) = 69x−2. Deduce the number of corn bags Bory can buy with 50,000 CFA.

Mathematics

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Answers

4 parts
a

25\dfrac{2}{5}

b

13\dfrac{1}{3}

3.2

13,5 cm213{,}5 \rm \, cm^{2}

3.3

sin⁡B=45,cos⁡B=35,tan⁡B=43\sin B = \dfrac{4}{5}, \quad \cos B = \dfrac{3}{5}, \quad \tan B = \dfrac{4}{3}

Step-by-step solution

15 steps

Part 3.1

Part a

25\dfrac{2}{5}

Calcul de P(A∪B)P(A \cup B)

  1. Formule d'inclusion-exclusion

    P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

  2. Substitution des valeurs

    P(A∪B)=0,2+0,3−0,1P(A \cup B) = 0{,}2 + 0{,}3 - 0{,}1

  3. Simplification

    P(A∪B)=0,4P(A \cup B) = 0{,}4

    ou en fraction

    P(A∪B)=15+310−110=210+310−110=410=25P(A \cup B) = \frac{1}{5} + \frac{3}{10} - \frac{1}{10} = \frac{2}{10} + \frac{3}{10} - \frac{1}{10} = \frac{4}{10} = \frac{2}{5}

    \dfrac{2}{5}

Part b

13\dfrac{1}{3}

Calcul de P(A∣B)P(A|B)

  1. Formule de probabilité conditionnelle

    P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

  2. Substitution

    P(A∣B)=0,10,3P(A|B) = \frac{0{,}1}{0{,}3}

  3. Simplification

    P(A∣B)=1/103/10=13P(A|B) = \frac{1/10}{3/10} = \frac{1}{3}

    \dfrac{1}{3}

Part c· 3 steps

Indépendance ?

  1. Condition d'indépendance

    P(A∩B)=?P(A)⋅P(B)P(A \cap B) \overset{?}{=} P(A) \cdot P(B)

  2. Calcul du produit

    P(A)⋅P(B)=0,2×0,3=0,06P(A) \cdot P(B) = 0{,}2 \times 0{,}3 = 0{,}06

  3. Comparaison

    0,06≠0,10{,}06 \neq 0{,}1

    Les événements A et B ne sont pas indépendants.

Part 3.2

13,5 cm213{,}5 \rm \, cm^{2}

Carré ABCD de côté 6 cm. M milieu de $[AB]$, N milieu de $[BC]$. Aire du quadrilatère AMNC (unités en cm²).

  1. Coordonnées (A en $(0,0)$, B$(6,0)$, C$(6,6)$, D$(0,6)$)

    M$(3,0)$, N$(6,3)$.

  2. Formule de l'aire d'un polygone (Schoelace)

    Points : A(0,0)(0,0), M(3,0)(3,0), N(6,3)(6,3), C(6,6)(6,6), A(0,0)(0,0)

    ∑xiyi+1=0⋅0+3⋅3+6⋅6+6⋅0=0+9+36+0=45\sum x_i y_{i+1} = 0\cdot0 + 3\cdot3 + 6\cdot6 + 6\cdot0 = 0 + 9 + 36 + 0 = 45 ∑yixi+1=0⋅3+0⋅6+3⋅6+6⋅0=0+0+18+0=18\sum y_i x_{i+1} = 0\cdot3 + 0\cdot6 + 3\cdot6 + 6\cdot0 = 0 + 0 + 18 + 0 = 18 Aire=∣45−18∣2=272=13,5 cm2{\rm Aire} = \frac{|45 - 18|}{2} = \frac{27}{2} = 13{,}5 \, \rm cm^2

    Vérification par intégration :
    De x=0x=0 à 3 : hauteur y=x−0y = x - 0

    ∫03x dx=[x22]03=92=4,5\int_0^3 x \, dx = \left[ \frac{x^2}{2} \right]_0^3 = \frac{9}{2} = 4{,}5

    De x=3x=3 à 6 : hauteur x−(x−3)=3x - (x-3) = 3

    ∫363 dx=3⋅3=9\int_3^6 3 \, dx = 3 \cdot 3 = 9

    Total : 4,5+9=13,5 cm24{,}5 + 9 = 13{,}5 \, \rm cm^2

    Autre méthode : aire carré 36 cm236 \, \rm cm^2 moins △MBN\triangle MBN.
    Base MB=3=3 cm, hauteur 3 cm :

    Aire △MBN=12⋅3⋅3=4,5 cm2{\rm Aire} \, \triangle MBN = \frac{1}{2} \cdot 3 \cdot 3 = 4{,}5 \, \rm cm^2

    Le reste n'est pas juste ce triangle (autre région), mais shoelace confirme.

    13{,}5 \rm , cm^{2}

Part 3.3

sin⁡B=45,cos⁡B=35,tan⁡B=43\sin B = \dfrac{4}{5}, \quad \cos B = \dfrac{3}{5}, \quad \tan B = \dfrac{4}{3}

△ABC\triangle ABC rectangle en A, AB =3= 3 cm, AC =4= 4 cm, BC =5= 5 cm. Calculer sin⁡B\sin B, cos⁡B\cos B, tan⁡B\tan B.

  1. Identification (angle en B)

    Côté opposé à B : AC =4= 4 cm
    Côté adjacent à B : AB =3= 3 cm
    Hypoténuse : BC =5= 5 cm

  2. sin⁡B\sin B

    sin⁡B=opposeˊhypoteˊnuse=45\sin B = \frac{\rm opposé}{\rm hypoténuse} = \frac{4}{5}

  3. cos⁡B\cos B

    cos⁡B=adjacenthypoteˊnuse=35\cos B = \frac{\rm adjacent}{\rm hypoténuse} = \frac{3}{5}

  4. tan⁡B\tan B

    tan⁡B=opposeˊadjacent=43\tan B = \frac{\rm opposé}{\rm adjacent} = \frac{4}{3}

    \sin B = \dfrac{4}{5}, \quad \cos B = \dfrac{3}{5}, \quad \tan B = \dfrac{4}{3}

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