6) \( \frac{3 a^{-2}-4 b^{-1}+a b}{3 a b-2 a} \quad a=3 ; b=4 \)

5 Completa a. \( (\ldots . . . \ldots . .+\ldots . .)^{2}=25 x^{2}+\ldots \ldots . . \ldots \ldots .20 x y+\ldots \ldots .-20 y \) b. \( \left(3 a^{2}+\ldots . .\right)^{3}=\ldots \ldots+54 a^{4} b+\ldots \ldots .+\ldots \ldots \) 6. Le dimensioni del quadro della signora Marilena sono \( x \) e \( y \) (in cm ). La cornice del quadro è uniformemente spessa 3 cm . Esprimi, in funzione delle dimensioni del quadro: a. l'area della tela (quadro senza cornice); b. l'area della cornice, osservando in particolare che essa dipende dal perimetro del quadro, non dalle sue dimensioni. 7. Un tuo compagno dice di aver scoperto che, moltiplicando per 10 un numero dispari, metà del numero pari che precede il numero dispari di partenza, si ottiene sempre un num

4. How many solutions does the equation \( 2(x+4)=2 x+8 \) have? A no solutions B. one solution C. two solutions D. infinite solutions

1. Find an odd natural number \( x \) such that \( \operatorname{LCM}(x, 40)=1400 \). 2. There are between 50 and 60 number of eggs in a basket. When Loza counts by 3 's, there are 2 eggs left over. When she counts by 5 's, there are 4 left over. How many eggs are there in the basket? 3. The GCF of two numbers is 3 and their LCM is 180 . If one of the numbers is 45 , then find the second number. 4. Using Mathematical Induction, prove the following: a) \( 6^{n}-1 \) is divisible by 5 , for \( n \geq 0 \). b) \( 2^{n} \leq(n+1)!, \quad \) for \( n \geq 0 \) c) \( x^{n}+y^{n} \) is divisible by \( x+y \) for odd natural number \( n \geq 1 \). d) \( 2+4+6+\cdots+2 n=n(n+1) \) e) \( 1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6} \) 47 f) \( 1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\frac{n^{2}(n+1)^{2}}{4} \) g) \( \frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1} \)

What does it mean for a number to be under a radical sign, and how does it affect the value of that number?

5.) \( \begin{array}{l}\text { If } f(x)=2 x+1 \text { sad } f(g(x))=6 x-3 \\ \text { then } g(0)= \\ \begin{array}{llll}\text { (A) }-2 & \text { (B) }-1 & \text { (C) } 0 & \text { (D) } 1\end{array}\end{array} \) (E) 3