<p>In an arithmetic sequence, a5 = a10 - 3, and a3 = -2. Between which two consecutive terms does 0 lie? </p>
<p>Can someone show how to do this problem?</p>
<p>also</p>
<p>What is the limit of (4x^2 + 10x + 4)/(3x + 6) as x approaches 2? I thought this problem was a piece of cake. I factored it to be (x+2)(X+.5) / 3 (x+2). Cancle out X+2 from top and bottom, Im left with x+.5 / 3. Plug in 2 and I guess 2.5/3 = .8333</p>
<p>However, When i graph the whole original equation and find the value at 2. I get get 3.3333</p>
<p>Why are these 2 answers different and why are they wrong? Thanks</p>
<p>just plug 2 into the original equation. u dont need to factor b/c 2 doesnt make the function undefined. dont do more work than u have to. and your factoring is wrong bc (x+2)(x+.5) = x^2+2.5X+1 which isnt 4x^2+10X+4</p>
<p>a5 = a1 + (5-1)b
a10 = a1+(10-1)b</p>
<p>a10 - a5 = [a1+(10-1)b] - [a1+(5-1)b] = 5b = 3</p>
<p>b = 3/5</p>
<p>a3 = a1+(3-1)b = a1 + (6/5) = -2</p>
<p>a1 = -16 / 5</p>
<p>To make it zero, we do</p>
<p>an = a1 + (n-1)b = -16/5 + (n-1)(3/5) = 0</p>
<p>n = 95 / 15. This says that Zero lies when n = 95/15, which is 6.3333, but n will always be an integer. Then, zero is somewhere between a6 and a7.</p>
<p>seven night, that is so weird! I've always used this trusty quadratic equation calculator and it gave me x=-2, -.5. So isnt that (x+2)(x+1/2)?</p>
<p>um after u get 5b how do you know to set it equal to 3?</p>
<p>ur program assumes the A in ax^2+bx+C will always be 1 (however the A in your equation is 4) this fx's the zeroes which the program doesnt take into account.</p>
<p>From the given information, it says a10 - 3 = a5, then, a10 - a5 = 3. a10 - a5 = 5b = 3.</p>
<p>nm got it man</p>
<p>How would one go about solving that limit problem if the denominator were (3x - 6)?</p>
<p>I've ggot two questions of my own, if you don't mind, johntam. :P I was actually going to make an identical thread.</p>
<p>When 4x² + 6x + L is divided by x + 1, the remainder is 2. Which of the following is the value of L?</p>
<p>(A) 4
(B) 6
(C) 10
(D) 12
(E) 15</p>
<p>And the other one...</p>
<p>If the ratio of sec(x) to csc(x) is 1:4, then the ratio of tan(x) to cot(x) is</p>
<p>(A) 1:16
(B) 1:4
(C) 1:1
(D) 4:1
(E) 16:1</p>
<p>ratio of tan(x) to cot(x) is 1/16 i believe.</p>
<p>Sorry. I meant to ask for an explanation; I have the answers, but don't know how to arrive at them.</p>
<p>You can do synthetic division</p>
<p>[-1]... 4.. 6.. L
.............-4.. -2
.........4.. 2.. 2</p>
<p>L+(-2)=2, so L=4</p>
<p>brinestorm...</p>
<p>first question:</p>
<p>if x+1 is a factor of F(x) or x = 1 is a zero of F(x), THEN F(1) = 0. However, if you evaluate the function at x=1, it actually gives a remainder. so F(1) = 4(-1)^2+6(-1)+L = 2. solve for L. answer should be (A) 4</p>
<p>second question.</p>
<p>sec/csc = (1/cos)/(1/sin) = sin/cos = 1/4</p>
<p>tan/cot = (sin/cos)/(cos/sin) = sin^2/cos^2 = (1/4)^2 = 1/16 or 1:16</p>
<p>don't do it by synthetic devision. what a waste of time. =D</p>
<p>how else tiger?</p>
<p>what what number were these questions cause i don't remember them on the test...but i didn't finish all the problems</p>
<p>...so that could be why i don't remember them (man i need some sleep...lol)</p>
<p><a href="mailto:ashernm@msn.com">ashernm@msn.com</a>, look at the my post above...</p>
<p>Re post #1 (alternate to post#4).</p>
<p>Useful fact:
for an arithmetic sequence am-an=(m-n)d.</p>
<p>a10-a5=3
a10-a5=5d
5d=3
d=3/5=.6.</p>
<p>a3=-2 is not too far from 0.
By hand or on a calculator
-2+.6 =-1.4=a4
+.6 =-.8=a5
+.6 =-.2=a6
+.6 =.4=a7
Zero is between a6 and a7.</p>
<p>Or a bit fancier:
[2ND] [STAT] [OPS] 5
seq(-2+.6X,X,0,5)
{-2 -1.4 -.8 -.2 .4 1} - this list contains 6 terms of the sequence starting with a3=-2.
Zero is between the 6th and 7th terms.</p>