This Blog exists for the collective benefit of all algebra students. While the posts are specific to Mr. Chamberlain's class, any and all "algebra-ticians" are welcome. The more specific your question (including your own attempts to answer it) the better.
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Thursday, April 14, 2011
hw #9-3 What goes up...
... must do homework!!
hw #9-3
pg 538-539 #26,27,40
pg 545 #26,34
pg 551 #9-33 Odd
I'm expecting questions on the blog... from EVERYONE!!
538#26 is the same as Karel dropping the rock off the bridge from 400ft... however this time it's an orange from 40 ft. Make a table of values and see what you get... you might end up using decimal values for t (time in seconds), such as 1.3, 1.4, etc...
Use the graphing calculator to graph the function (using 'x' instead of 't')... does your graph 'agree' with your table of values?
Experiment, explore... this is the fun stuff!
... and yes, you guessed it, we will soon be exploring more algebraic (i.e. more fun) ways to solve a problem like this!!
If you plugged in 0 as your height and then solved for t would that give you your impact time for "dropping problems" (i.e. karel's rock problem). If it works, that would be a much easier way to solve those problems than plugging in a table of values. That takes FOREVER!!!!!!!! It also wastes paper. Half of my page was the table of values!
JULIA SPOILS ALL THE FUN!! Yes Julia, you are correct, since we are often interested in the amount of time it takes for an object to hit the ground, well gosh!... h is the height at any given time, so we just want to know the value of 't' when h=0.
For #26, Yes, eukinjus (ah-hah!!, the eukinjus (i.e. "you can just") monster rears its ugly head) graph the "right side" of the parabola, since the domain for this real-life function exists only for non-negative values of t. Of course, if you graph the right side, eukinjus graph the left side using symmetry, yes?
#40 was a 2-part problem... the first part asked you to use whole numbers for 't' and estimate... the second part simply suggests that eukinjus use decimal values, such as 3.1,3.2,etc. to make your estimate more precise.
btw... (perhaps you recall) as we were ending class the other day, I was trying to show what would happen if we vertically propelled a rock 400ft into the air and then awaited its return to terrafirma (look it up).
The equation for Karel was h=-16t^2+400, since there was no initial velocity and the initial height was 400ft. PaulW astutely pointed out that the "terminal velocity" was -160 feet/sec, based on 5 seconds worth of increased gravitational pull.
EXTRA CREDIT: Go to "graphit" and graph the function h=-16t^2+160t+0. Compare (graph both with a comma to separate) and explain FULLY what is happening in the two graphs. What is similar? What is different? Why?
N.B. remember on graphit, you have to use 'x' instead of 't'
Square roots of positive integers are REAL, though many times not RATIONAL.
Only perfect squares, e.g. 1,4,9,16,25,... yield integer square roots. Some rational numbers have rational roots, e.g. sqrt of 6.25 is 2.5, (notice that the sqrt of 625 is 25, see the resemblance?)
The square roots of all positive integers that are not perfect squares are irrational, e.g. sgrt of 5, 6,7,8,10,11,...etc.
We will be reviewing square roots, Julia's solution, and more in class!! Betcha can't wait??!!
EC: The Graphs are similar because they are the same size and shape. They are different because one represents a Vertical Motion equation while the other represents a Vertical drop equation. They do because the first equation, -16x^2+400, has no v or velocity which is given when an object is thrown. The second equation, -16x^2+160x+0 represents the vertical motion because it involves a velocity of 160 ft/sec.
My questions will come on Sunday..
ReplyDeleteConfused on number 26 page 538
ReplyDelete538#26 is the same as Karel dropping the rock off the bridge from 400ft... however this time it's an orange from 40 ft. Make a table of values and see what you get... you might end up using decimal values for t (time in seconds), such as 1.3, 1.4, etc...
ReplyDeleteUse the graphing calculator to graph the function (using 'x' instead of 't')... does your graph 'agree' with your table of values?
Experiment, explore... this is the fun stuff!
... and yes, you guessed it, we will soon be exploring more algebraic (i.e. more fun) ways to solve a problem like this!!
If you plugged in 0 as your height and then solved for t would that give you your impact time for "dropping problems" (i.e. karel's rock problem). If it works, that would be a much easier way to solve those problems than plugging in a table of values. That takes FOREVER!!!!!!!! It also wastes paper. Half of my page was the table of values!
ReplyDeleteFor #26, do we only need to graph the right side of the parabola?
ReplyDeleteBy the way, I used up a whole paper doing that one problem!
If I changed the window's scale on the Graphit to what I used on my paper & the graph looked the same, is it correct?
ReplyDeleteWhat is the answer to #26?
ReplyDeleteIt's not in the back of the book.
I graphed my parabola and got that the orange hits the water at about 1.6 seconds..
I just forgot, does the coefficient of 't' in
ReplyDelete-16t^2+c affect the way the parabola opens? Or what does?
For #40, isn't the answer kind of obvious, for the first bullet, plug in values for 't'?
ReplyDeleteWhat does #40 mean by "How can you use a second table to make your approximation more accurate?"
ReplyDeleteFor all of the "related function" questions like #9, do you just switch out the 0 and put in a y?
ReplyDeleteJULIA SPOILS ALL THE FUN!! Yes Julia, you are correct, since we are often interested in the amount of time it takes for an object to hit the ground, well gosh!... h is the height at any given time, so we just want to know the value of 't' when h=0.
ReplyDeleteFor #26, Yes, eukinjus (ah-hah!!, the eukinjus (i.e. "you can just") monster rears its ugly head) graph the "right side" of the parabola, since the domain for this real-life function exists only for non-negative values of t. Of course, if you graph the right side, eukinjus graph the left side using symmetry, yes?
#40 was a 2-part problem... the first part asked you to use whole numbers for 't' and estimate... the second part simply suggests that eukinjus use decimal values, such as 3.1,3.2,etc. to make your estimate more precise.
ReplyDeletebtw... (perhaps you recall) as we were ending class the other day, I was trying to show what would happen if we vertically propelled a rock 400ft into the air and then awaited its return to terrafirma (look it up).
ReplyDeleteThe equation for Karel was h=-16t^2+400, since there was no initial velocity and the initial height was 400ft. PaulW astutely pointed out that the "terminal velocity" was -160 feet/sec, based on 5 seconds worth of increased gravitational pull.
EXTRA CREDIT: Go to "graphit" and graph the function h=-16t^2+160t+0. Compare (graph both with a comma to separate) and explain FULLY what is happening in the two graphs. What is similar? What is different? Why?
N.B. remember on graphit, you have to use 'x' instead of 't'
What do we compare h=-16t^2+160x+0 to?
ReplyDelete... and when is Julian ever going to collect that beanie-baby??
ReplyDeleteWait-- What do we use as 'x' in h=-16t^2+160x+0?
ReplyDeleteh=y, but if we change t to x, then what do we do to x in that equation?
Dear #26 graphers/workers/paper pushers... I assume that you will be paying attention when Julia presents her two-step solution?
ReplyDeleteI don't really understand what Julia means by that solution.. Will she explain it during class?
ReplyDelete0 is not a real number, right?
ReplyDeleteI think any number other than the square root of a negative is real, so 0 is real.
ReplyDeleteSquare roots of positive integers are REAL, though many times not RATIONAL.
ReplyDeleteOnly perfect squares, e.g. 1,4,9,16,25,... yield integer square roots. Some rational numbers have rational roots, e.g. sqrt of 6.25 is 2.5, (notice that the sqrt of 625 is 25, see the resemblance?)
The square roots of all positive integers that are not perfect squares are irrational, e.g. sgrt of 5, 6,7,8,10,11,...etc.
We will be reviewing square roots, Julia's solution, and more in class!! Betcha can't wait??!!
Zero is a very real number. Maybe we should have a debate. State the affirmative case!
ReplyDeleteAccording to the above, does 0 have a square root of 0?
ReplyDeleteWell, what is zero*zero?
ReplyDeleteZEROOOOO!!
ReplyDeleteFor EC aren't those just the same graph?
ReplyDeleteEverything YOU need to know about REAL NUMBERS...
ReplyDeletehttp://www.jamesbrennan.org/algebra/numbers/real_number_system.htm
Dear EC,
ReplyDeletef(x)=-16x^2+400
g(x)=-16x^2+160x+0
Look at your class notes as to just what a,b,&c represent.
EC: The Graphs are similar because they are the same size and shape. They are different because one represents a Vertical Motion equation while the other represents a Vertical drop equation. They do because the first equation, -16x^2+400, has no v or velocity which is given when an object is thrown. The second equation, -16x^2+160x+0 represents the vertical motion because it involves a velocity of 160 ft/sec.
ReplyDelete