<p>1) IF (A^3)=(B^2) WHICH OF THE FOLLOWED STATEMENTS COULD BE TRUE?</p>
<p>A)A<0 AND B>0
B)A>0 AND B<0
C) A>0 AND B>0</p>
<p>A) A ONLY
B) B ONLY
C) C ONLY
D) A AND B
E) B AND C</p>
<p>2) 3y=2X<em>6
Y=1</em>CX
IN THE EQUATION ABOVE,X AND Y ARE VARIABLES AND C IS CONSTANT. IF NO ORDERED PAIR OF NUMBERS (X,Y) SATIFIES BOTH OF THE EQUATONS ABOVE,WHAT IS THE VALUE OF C??
A)-1.5
B)(-2/3)
C) 0
D) (2/3)
E) 3/2</p>
<h1>1: E; if a was negative, a^3 would be negative. Since anything squared is always positive, “A” cannot be true.</h1>
<p>If a=1, then a^3=1. If b=-1, then b^2=1. Therefore a^3=b^2</p>
<p>If a=1, then a^3=1. If b=1, then b^2=1. Therefore a^3=b^2</p>
<p>Thank u
what about the second question??</p>
<p>The lines must be parallel to have no common x,y coordinates. IOW, they must have the same slope. Set both equations to y = (m)x + k format. What does the constant (C) have to be to make the slope (m) equal in both equations?</p>
<p>3y = 2x -6 becomes y = (2/3)x - 2</p>
<p>y = 1 - Cx becomes y = -Cx + 1</p>
<p>In order for the slopes to be equal in both equations C must equal -(2/3)</p>
<p>I’ll assume that the underscore in your post is a minus sign.</p>
<p>So the two equations are: 3y=2x-6, and y=1-Cx. Solve these by substituting 1-Cx for y in the first equation. You get 3-3Cx=2x-6. Solve for x. You get x=9/(2+3C).</p>
<p>For what value of C is x not a “specific” number? Well what happens when C=-2/3? Then x in not defined. Some may say it is infinite. However you think of it, the two equations don’t then have a solution. So the choice is (B).</p>
<p>All this said this doesn’t seem like a particularly well posed question. Did it actually appear on some previous SAT test?</p>
<p>Thank u all
fogcity yes it is from a real sat test</p>