SAT PRACTICE TEST 2: SECTION 1

IF  $Y=\sqrt{X},\ THEN\ 2+Y^2=$  

• 2

• 2X

•  $2+\sqrt{X}$  

• 2+X

•  $2+X^2$  

IN THE EQUATION ABOVE, A=5, B=7, AND  $B^2-4AC=0$ . WHAT IS THE VALUE OF X?

•  $-\frac{7}{5}$  

•  $-\frac{7}{10}$  

• 0

•  $\frac{7}{10}$  

•  $\frac{7}{5}$  

IF  $2N^2\ IS\ AN\ EVEN\ INTEGER$ , WHAT IS TRUE ABOUT  $N+1$  ?

• IT IS AN EVEN INTEGER

• IT IS AN ODD INTEGER

• IT CAN BE AN ODD OR EVEN INTEGER

• IT IS A PRIME NUMBER

• IT IS A PERFECT SQUARE

WHAT IS THE SLOPE OF LINE N IF  $LINE\ N\perp\ LINE\ M$  ?

•  $-\frac{3}{2}$  

•  $-\frac{2}{3}$  

•  $\frac{2}{3}$  

•  $\frac{3}{2}$  

• CANNOT BE DETERMINED FROM THE INFORMATION PROVIDED.

IF THE SOLUTION SET OF THE INEQUALITY ABOVE WERE GRAPHED ON A SET OF COORDINATE AXES, THE GRAPH WOULD BE A

• A SOLID VERTICAL LINE THROUGH $X=-2$ , SHADED TO THE LEFT.

• SOLID VERTICAL LINE THROUGH $X=-2$ , SHADED TO THE RIGHT.

• DOTTED VERTICAL LINE THROUGH $X=-2$ , SHADED TO THE LEFT.

• DOTTED VERTICAL LINE THROUGH $X=-2$ , SHADED TO THE RIGHT.

• DOTTED HORIZONTAL LINE THROUGH $X=-2$ , SHADED ABOVE THE LINE.

IF THE LINE SHOWN ABOVE IS REFLECTED OVER THE X-AXIS, WHAT IS THE EQUATION OF THE NEW LINE?

• $Y=-\frac{2}{3}X+4$

• $Y=\frac{2}{3}X-4$

• $Y=-\frac{2}{3}X-4$

• $Y=\frac{2}{3}X+4$

• $Y=-\frac{3}{2}X+4$

WHICH POINT SATISFIES THE SYSTEM OF LINEAR INEQUALITIES?

• (-1,-2)

• (-1,2)

• (1,2)

• (2,-3)

• (2,3)