<p>Hey guys-- i was trying to solve this problem with long division by testing all the possible k values given in the choices, and picking the k value which doesn't result in any remainder.but i'm wondering if theres an easier way to go about it without testing all five numbers.</p>
<p>*if x-2 is a factor of x^3 + kx^2+ 12x - 8, then k= *</p>
<p>a) -6
b) -3
c) 2
d) 3
e) 6</p>
<p>Yes. Its called synthetic divison. Heres how it works. You make a table of the coeffiecients to start
2 l 1 k 12 -8
l
l________________</p>
<p>and you have the solution to (x-2) = 0 [for any factor] to the left of the table, now you bring down the 1 and then multiply it by 2 like so:</p>
<p>2 l 1 k 12 -8
l \l/ 2
l_________________
1 1*2 =2 (\l/ is an arrow fyi)</p>
<p>then you add the next two terms and multiply the answer again by 2</p>
<p>2 l 1 k 12 -8
l \l/ 2 2k+4
l__________________
1 k+2 2(k+2) = 2k+4 </p>
<p>now we continue to do this until the last term</p>
<p>2 l 1 k 12 -8
l \l/ 2 2k +4 4k+32
l________________________
1 k+2 2k+16 4k+24 you get the idea</p>
<p>Now the point of this is that in oder for x-2 to be a factor of the function, the last thing we get w/ synthetic division must equal 0.</p>
<p>4k + 24 = 0
4k = -24
k = -6</p>
<p>Horray!!!!!!! the answer is A!</p>
<p>thanks a lot-completely forgot about synthetic division. We should ask the administrator of this board to add a 'math symbols' feature. Makes it a lot easier to type out solutions to math problems. Mind helping me out with this as well?--</p>
<p>** A line has parametric equations x=5+t and y= 7+t where t is the parameter. The slope of the line is**</p>
<p>a)5/7
b) 1
c) 7+ t / 5 + t
d) 7/5
e) 7</p>
<p>System of equations?</p>
<p>x = 5 + t
t = x - 5</p>
<p>y = 7 + t
y = 7 + x - 5
y = x + 2</p>
<p>Slope would be 1.</p>
<p>It's 1...if you graph it and then use the table</p>
<p>thanks-- i feel so silly-- simple y= mx + C, where m is the slope...</p>
<p>check this one out</p>
<p>The number of hours of daylight, d , in Hartsville can be modeled by
d = 35/3 + 7/3 sin (2.pi.t / 365), where t is the number of days after March 21. the day with the greatest number of hours of daylight has how many **more daylight hours than May 1? ( March has 31 days, April has 30 days,May has 31 days, June has 30 days)**</p>
<p>a) 0.8 hr
b) 1.5 hr
c) 2.3 hr
d) 3.0 hr
e) 4.7 hr</p>
<p>What i did was i plugged in 41 for t( number of days from march 22 to may 1). Then i plugged in 92 for t(number of days from march 22 to june 21). Then i subtracted the result of the former from the result of the latter. But i didn't get anything even remotely close to any of the answer choices.</p>
<p>To your first one, it can be even easier. Plug in x=2 and solve for k, setting the equation to 0. :)</p>
<p>It becomes:
8+4k+24-8=0
4k=-24
k=-6</p>
<p>Same with the second. Pick:
t=0: (5,7)
t=1: (6,8)</p>
<p>Slope: 1</p>
<p>As for the third, I don't like it. I wouldn't have known which day had the most daytime hours. :P</p>
<p>Was the answer to the third one C) 2.3?</p>
<p>arthurbulla- i know right? WHY do they expect students to know this? it's a MATH test. I VAGUELY remember, from my environmental science class, that the summer solstice(sp) is on june 21st- which is supposed to be the hottest day of the year.</p>
<p>specify- i looked up the answer and its 0.8hrs. They dont even explain how they got that answer. So if anyone knows- help out----</p>
<p>On May 1st, t = (31-21) + 30 + 1 = 41.</p>
<p>On May 1st, d(41) = 35/3 + 7/3 sin (2.pi.(41) / 365)
On longest day, d(?) = 35/3 + 7/3 sin (pi/2) [since d is maximized when sin() hits a max, which is at pi/2]</p>
<p>Diff: d(?) - d(41) = 7/3( sin(pi/2) - sin( 82 pi/365))
= (7/3) ( 1 - sin( 0.7058 radians))
= (7/3) (1 - 0.64864)
= 0.82 hours</p>
<p>Looks like answer (a) .</p>
<p>(I simplified a bit above. Since t can have only integer values, it's unlikely that 2.pi.t / 365 will <em>exactly</em> equal pi/2, on the 'longest' day. However, since you only need an approximate answer to pick the 'best fit' of the 5 possible answers, I think it's a reasonable simplification.)</p>
<p>THANKS!- cant believe i didn't think about that- the trig function is the only thing that changes the value of d so to find the greatest value of d, the greatest possible value of sin() has to be found. So all i had to do was equal ** 2.pi.t/365** to pi/2 and solve for t. Then plug that into the whole equation, and find d. And then subtract the d value obtained by plugging in 41 from the d value obtained by using pi/2.</p>
<p>caa5042:</p>
<p>You went one better. You're right - solve for t from (2.pi.t/365) = pi/2, and you will get t = 365/4 = 91.25 . Round this off to the nearest integer value, and you get t=91. </p>
<pre><code>Now just compute d(91) - d(41) , and you are all set....
</code></pre>
<p>cant you just graph and subtract? just a suggestion, hopefully you have a graphing calculator</p>