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#1 |
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(づ ̄ ³ ̄)づ watermelon
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THIS THREAD IS USELESS NOW UNLESS YOU SOMEHOW HAVE THE SAME QUESTIONS AS THESE.
I thank all the people who attempted and took time to help me out. <33 You guys are the best. ^^--- M'kay. So my teacher Mr. Furry (his real name, I'm not kidding) just gave my class and I some questions that we didn't even learn. =/ It's due tonight at 11:55 PM EST... so, if you have some time (or if you are extremely bored), please help! I have NO idea how to do these. Lol. Any help is appreciated. (I'm in high school, btw.) Thanks in advance. Here are the questions that I don't know how to do: 1. The area of a square with sides of length s is given by A = s2. Find the rate of change of the area with respect to s when s = 5 meters. Answer is in meters^2. And it is not 25. 2. Consider the following function. f(t) = 6t^2 - 4 (a) Find the average rate of change of the function below over the interval [5, 5.1]. Answer is not the average of the numbers plugged in... (b) Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. Answer will have one at t=5 and another with t=5.1 3. Use the position function -16t^2 + vot + so for free-falling objects. (a) A ball is thrown straight down from the top of a 240 foot building with an inital velocity of -25 feet per second. What is its velocity after 5 seconds? Answer is in ft/sec. (b) What is its velocity after falling 103 feet? Answer is also in ft/sec. 4. Find k such that the line y = 5x + 6 is tangent to the graph of the function below. y=k*sqrt(x) Answer is what k equals. 5. The tangent line to the graph of y = h(x) at the point (-7, 6) that passes through the point (2, 8). Find the following values. (a) h(-7) (b) h '(-7) 6. (Use the position function -4.9t^2 + vot + so for free-falling objects.) To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen 5.2 seconds after the stone is dropped? -- Edit Answer for 1 is 10. -- Edit #2 Answer is 60.6, 60, and 61.2 respectfully. -- Edit #3 Disregard #3. I maxed out my submissions for the question. xD 3a was -185 though. -- Edit #4 Answer is 11. I BSed that one. -- Edit #5 Answer to part 2 is 2/9. Maxed out submissions for part 1, so disregard this question now. And... YAY ONLY #6 LEFT! -- Edit #6 Answer is 132.496. ![]()
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Last edited by Windscarredfaith; 12-7-2006 at 08:49 PM.. |
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#2 |
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is against custom titles
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1. dA/ds = 2s. dA/ds|s=5 = 2(5) = 10, though I don't see how it can be in m^2
2. df/dt = 12t. df/dt|t=5 = 60. df/dt|t=5.1 = 61.2 Average = 60.6 3. s = -16t^2 + vot + so v = -32t + vo v|vo=-25, t=5 = -160 - 25 = -180 (b) s = -16t^2 + vot + so 137 = -16t^2 - 25t + 240 EDIT: d'oh about 3 --Guido http://andy.mikee385.com
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![]() Last edited by GuidoHunter; 12-7-2006 at 08:39 PM.. |
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#3 |
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(づ ̄ ³ ̄)づ watermelon
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Lol, yeah... those are the questions I got.. ^^"
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#4 |
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Custom User Title
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On #5, find the y=mx+b for the tangent line. Slope is the difference in y over difference in x. Substitute an x and y pair in to find b. Then with h(x), plug the x value into your new mx+b to find y.
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#5 |
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Falcon Paaaauuuunch!!!!!!
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Here's what I got for 5, haven't done the others yet.
5. tangent line -> y=9/2x +1, so h(7)=9/2(7) +1 = 22.5, and h'(7)=9/2
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#6 |
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(づ ̄ ³ ̄)づ watermelon
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Tried the answers. Both wrong. =/
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#7 |
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Custom User Title
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The slope on that should be 2/9 not 9/2.
=> h'(x) = 2/9
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#8 |
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Resident Penguin
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3a is -205 but I got that doing it in my head so it's probably wrong.
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#9 |
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(づ ̄ ³ ̄)づ watermelon
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Yeah. That one is right. <-- to what itmorr said
And, disregard 3, my submissions are gone.
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#10 |
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Falcon Paaaauuuunch!!!!!!
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Okay, here's another, although very confusing of a way of doing it, but I think it's the right answer.
6. x=xo +vot +1/2at^2, so x-xo=1/2*4.9*5.2^2=66.248meters
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#11 |
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Resident Penguin
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well 6 is easy just move the h sub zero over to the left side and plug in the 5.2 and I think that will do it.
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#12 |
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(づ ̄ ³ ̄)づ watermelon
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Nope.... Lol. Sorry. ^^"
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#13 |
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Falcon Paaaauuuunch!!!!!!
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Tried 5 again.
5. tangent line -> y=2/9x +68/9, so h(7)=2/9(7) +68/9 = 89/9, and h'(7)=2/9 Even though you're done with that now. Edit, okay, so my answer for 6 was wrong, because the -4.9 already had the 1/2 in it, so I now think it's 132.496meters.
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#14 |
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(づ ̄ ³ ̄)づ watermelon
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Lol. Thanks for that. I got 1/2 credit. xD
-- Edit YAY! #6 is right. ^^
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Last edited by Windscarredfaith; 12-7-2006 at 08:48 PM.. |
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#15 |
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Custom User Title
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Does this work on the one you BS'd as 11?
y=k*sqrt(x) curve y = 5x + 6 tangent to curve 5x + 6 = k*sqrt(x) 5x + 6 over sqrt(x) = k
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#16 |
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(づ ̄ ³ ̄)づ watermelon
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Thanks for the explanations, itmorr. I'll definitely try to teach myself now since our teacher doesn't. =/
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#17 |
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Falcon Paaaauuuunch!!!!!!
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This isn't useless unless someone has the same questions as these, people can use this to find out how to do similar problems.
I'm glad I could help.
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#18 |
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Super Scooter Happy
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How the hell is #6 a Calculus question? That's Algebra II at worst.
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I watched clouds awobbly from the floor o' that kayak. Souls cross ages like clouds cross skies, an' tho' a cloud's shape nor hue nor size don't stay the same, it's still a cloud an' so is a soul. Who can say where the cloud's blowed from or who the soul'll be 'morrow? Only Sonmi the east an' the west an' the compass an' the atlas, yay, only the atlas o' clouds. |
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#19 |
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(づ ̄ ³ ̄)づ watermelon
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^^" I dunno. Lol. Maybe because it's high school Calculus?? It's a senior course, but I'm taking it as a Junior because I skipped a level. I feel left out there. xD
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#20 |
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Super Scooter Happy
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I did partial fractions in high school calc. That's not an excuse. >_>
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I watched clouds awobbly from the floor o' that kayak. Souls cross ages like clouds cross skies, an' tho' a cloud's shape nor hue nor size don't stay the same, it's still a cloud an' so is a soul. Who can say where the cloud's blowed from or who the soul'll be 'morrow? Only Sonmi the east an' the west an' the compass an' the atlas, yay, only the atlas o' clouds. |
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