<p>if x is a real number, which of the following is a graph of the solution set of (x-3)(2-x) greater than or equal to zero?
just explain how the number line would be</p>
<p>for how many different values of x is (x^2-x-6)^(x^2 -3x+2) equal to 1 ?</p>
<p>heyyyyyyyyy anyone ??</p>
<p>Here are some hints:</p>
<ol>
<li><p>The roots of the equation are 2 and 3. Do casework based on whether x < 2, 2 < x < 3, and x > 3 (you can do this by plugging in numbers and checking whether the expression is positive or negative). This should give you the correct interval.</p></li>
<li><p>For real x, (x^2 - x - 6)^(x^2 - 3x + 2) = 1 only when</p></li>
</ol>
<ul>
<li>x^2 - x - 6 = 1</li>
<li>x^2 - x - 6 = -1 and x^2 - 3x + 2 is an even integer</li>
<li>x^2 - 3x + 2 = 0 and x^2 - x - 6 is nonzero (otherwise 0^0 is indeterminate).</li>
</ul>
<p>Check each of these three cases to find all solutions.</p>
<p>ummm…okay i didnt get the 1st one i got the inerval x < 2, and x > 3 how did u get 2 < x < 3 !!!</p>