1T 2010 høst LØSNING

DEL 1.

Oppgave 1

a)

<tex> x+y=4 \wedge 3x - y =8</tex>

<tex>y=4-x \wedge 3x-4+x=8</tex>

<tex>y=4-x \wedge 4x=12</tex>

<tex>y=4-x \wedge x=3</tex>

<tex>y=1 \wedge x=3</tex>

<tex>x=3 \wedge y=1</tex>

b)

<tex>- \frac14x+2 =2x- \frac52</tex>

<tex>x+8 =8x- 10</tex>

<tex>- 9x = - 18</tex>

<tex>x = 2</tex>

c)

<tex>5,7 \cdot 10^4 + 3,0 \cdot 10^3 = 57000 + 3000 = 60000 = 6,0 \cdot 10^4</tex>

d)

<tex> \frac{3}{x+4} + \frac{24}{x^2-16} = \frac{3}{x+4} + \frac{24}{(x+4)(x-4)}= \frac{3 (x-4)}{(x+4)(x-4)} + \frac{24}{(x+4)(x-4)}= \\ \frac{3x-12}{(x+4)(x-4)} + \frac{24}{(x+4)(x-4)}= \frac{3x + 12}{(x+4)(x-4)} = \frac{3(x+4)}{(x+4)(x-4)}= \frac{3}{x-4} </tex> e)

f)

g)

1)Sannsynlighet for å like begge:

<tex>P(liker \quad begge) =\frac{16}{25}\cdot \frac{15}{24} = \frac 35</tex>

Oppgave 2

DEL 2