<p>I'm struggling with these questions. Thanks for the help.</p>
<p>yfrog</a> Photo : Shared by</p>
<p>yfrog</a> Photo : Shared by</p>
<p>yfrog</a> Photo : Shared by</p>
<p>2:</p>
<p>(2n+1)(3n-11) < 0
Easiest way, I think, would be to plug it into your calculator
so y1 = (2x+1)(3x-11)
Then go to the table, and look for negatives.</p>
<p>You find 0,1,2,and 3 give negative numbers.So (E), four.</p>
<p>3:
The distance from each must be the same so:
x - 3 = y - x = (2x - y) / 2</p>
<p>first two:
x -3 = y - x</p>
<p>Solved for one variable:
2x = y + 3</p>
<p>now you can substitute it in the last one</p>
<p>(y + 3 -y) / 2 = 1.5</p>
<p>Thus the distance from each is 1.5</p>
<p>so x = 3, next 4.5, then y is 6</p>
<p>1: First, you can discount answers d and e because those aren’t related to the height of the stack.</p>
<p>Next, find c. If stack1 is 4 in taller than stack2, and has 2 more buckets, then the part that sticks out must be 2 in. -seems like c could be answer</p>
<p>Next, find a, height of one bucket. stack2 is 14 in tall. 14-4=10. A is 10</p>
<p>Next find b, the overlap 10-2=8</p>
<p>So, if all buckets add 2 in to the height, and one adds another 8 in, then the 2 in 2n+8 must be variable c.</p>
<p>ur right but i was wondering if there was a more straight forward way in the second question because i don’t like this back solving method</p>
<p>wittynickname, ur right.</p>
<p>For question 2:
First, because it is a product, the answer will only be negative if one term is positive, and one is negative. </p>
<p>Find when the first term is negative
(2n+1)<0 2n<-1 n<-1/2 so integers that make term1 negative are -1,-2 etc.</p>
<p>Find when the second term is negative
note that because you subtract 11, any negative n will give a negative answer
(3n-11)<0 3n<11 n<3.6
so integers are 3,2,1,0,-1 etc.</p>
<p>now choose numbers that only make one term negative(appears in only one solution set)</p>
<p>so the only integers that make one term negative and the other positive are 0,1,2,3</p>
<p>Here’s an algebraic solution to question 2:</p>
<p>First figure out the real numbers where the expression is 0. To do this just set each factor to 0, and you get -1/2 and 11/3. </p>
<p>If you had to take an educated guess as to where the negative solutions were, then I think it seems reasonable to pick all the integers between these two numbers, ie. 0, 1, 2, and 3. Thus there are four solutions.</p>
<p>Let me first give a remark, then I’ll give some deeper mathematics:</p>
<p>Remark: We choose the “inside” of the two numbers because the answer choices indicate that there must be finitely many - the “outsides” give infinitely many.</p>
<p>Deeper understanding: The given expression is a polynomial (in particular, it is quadratic, so that its graph is a parabola). Polynomials are “continuous everywhere,” that is you never lift your pencil from your paper when drawing them. Thus, the only way they can change from positive to negative (or vice versa) is by passing through zero. So once you find the zeros, you need only test one value in each of the intervals determined by these zeros (in this case there are three) to determine if you get negative or positive results in the WHOLE interval.</p>
<p>I would recommend doing the third one by “starting with choice (C).” In other words substitute each answer choice in for y and see if it works. If you want me to write out the details of this method, let me know. I just don’t have time to do it right now.</p>
<p>Thanks, DrSteve. No, I think i understand now.</p>