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we can formulate a "slope" with the ratios provided I believe
45/2 143/4 869/10 might help ....
k2x = 45 k4x = 143 k10x = 869 ... or maybe plot the points on a graph and find a quadratic that matches the points?
does it come with a quadratic function for us to use?
i could find out but i would have to close the assignment to open another and it may not come up with the same question
sounds like mathLab...
it's through plato. i do all online courses for school.
if it doesnt provide a quadratic to begin with, you probably have to use this data to construct one.....
A(2^2) +B(2) +C = 45 A(4^2) +B(4) +C = 143 something like that
The ABC that are normally constants become the "variables" to find... in a system of equations
thanks but someone already helped me figure it out (:
yay!! ... howd they do it?
i really don't know. they just gave me the equation and what to plug in and that gave me the answer. now 4 more questions to go!! After a second, the rock is 58 feet in the air; after two seconds , it is 112 feet in the air. find the height, in feet, of the rock after 10 seconds.
ahhh... instead of teaching youto fish, they simply gave you a fish....
the rock question usually comes with a formula for the effect of gravity on an object....
is this a physics quiz or a math quiz? the difference being of course that physics wants youto apply a formula, and math wants you to figure out the formula...
it's algebra 2
Newtons here, hes gravity based I think :)
I answered it, and for the record, 'amistre64' I tried for a considerable amount of time to give hints without the answer (and a full method is there anyway, if someone has time to read my entire posts...)
Newton, kudos then :) I recant my fishy commentary :)
I'll carry on the answer to the gravity one in here then, rather than clog up this place; for the record, are we taking g = 10 or 9.8 or what?
what do you mean
I believe it it impossible to find an answer if the solution is from a general quadratic Distance = ax^2 + bx + c (but if I am wrong someone say), so we need to simplify it. either we assume it started on the ground (c=0), or assume the value of g.
it says "rock launched from cannon"
OK, if we assumed it was from the ground, we can use the equation: s = ut - (1/2)at^2, or in this case height = (initial speed)*t - (g/2)t^2 Using this we can form two equations from the given information: 58 = u(1) - g/2(1^2) => 58 = u - g/2 112 = (u)2 - g/2(2^2) => 112 = 2u - 2g We can find g and u from this, and use them in the eqation: height = u(10) - (g/2)(10^2) for time = 10. Try it
This gives a wildy inaccurate value of g, though, so I am unsure if it was the intention. See if you can get an answer and check if it is what they wanted; if not, post back.
i got 544 but it was wrong
next queston; the company discovered that it costs $45 to produce 3 camera cases, $257 to produce 7 camera cases, and $792 to produce 12 camera cases. find the total of one camera case. use the quadratic function to solve.
i think the answer would be 15 for this question
Try 400? If it is right, I will give a much more detailed explanation; if it was wrong I will think of a new plan (if you can still try the rock one)
it only gives me one chance & i have to move on to the next question
Oh OK then. So why do you think this one is 15? What is your method?
Well, I don't think it is 15 (so I wouldn't enter that!) For this one, you know it is controlled by a general quadratic function, so let this be cost = ax^2 + bx + c (x = number of cameras) You can generate equations using the given information: 45 = (3^2)a+3b+c (*) 257 = (7^2)a+7b+c (**) 792=(12^2)a+12b+c (***) You can solve these equations to give values for a, b and c, and then use cost = ax^2 + bx + c with x = 1 to find the cost of 1 camera.
how do i do that
To start off: (**) - (*) gives 257 - 45 = (49 - 9)a + (7-3)b => 212 = 40a + 4b (***) - (**) gives: 792 - 257 = (144-49)a + (12-7)b => 535 = 95a+5b You now have: 212 = 40a + 4b 535 = 95a+5b Can you solve these two for a and b? (Then plug these back into (*) for c)
i'm lost on how to got it to this point or how to solve for a and b.
OK - do you see how the question can be used to get the first 3 equations with a, b and c, though? Or not that, either?
if i was to use a graphing calculator, how do i figure it out?
OK. I do not use graphics calculators, but I assume if you plot the points (x,y): (3,45); (7, 257); (12, 792) Then it can solve it for y = f(x) = ax^2 + bx + c, with values for a, b and c, which you can use to find f(1). But really, what I did above is just solving simultaneous equations, which will give you the same answer: Take one away from the other to eliminate c; do this again; get two equations in only a and b. Then use these to eliminate either a or b (probably b), etc.
i was just seeing which way would be easier and less confusing
i was just seeing which way would be easier and less confusing
less confusing? i didnt think that was possible in math :) but I like newtons way, it just validiates what I was thinking as a solution in me head
i mean when i see what he is doing, i think i understand but then when it comes to doing it on my own, i get lost /:
how good are you at "solving a system of equations"?
either by elimination, substitution, or matrix.... they all amount to the same thing in the end
Yes, as above. Although, personally, I like to avoid matrices unless absolutely necessary!
Matrices can get rather confused if you dont keep track of all your work...so on complicated stuff, id perfer the elimination/substitution as well.
i mean if it was down to where you plug in the annswer, i could do it. but figuring it out to there is completely confusing to me.
9a+3b+c = 45 49a+7b+c = 257 144a+12b+c = 792 What do you "feel" is the first step to solving this system of equations? lets step through it
9a+3b+c = 45 to solve this
ok ... but we can only solve it part way... because we have other variables to consider; but would you agree that: c = 45 -9a -3b?
we can use this "value" of c to work the other 2 equations with... but first we have to agree that this is a good "value" for c
I think that is probably a better way to go about it if you are explaining how to do it along the way than mine, actually; regardless, they appear to have left us. Ah well.
mighta got booted :) theyll be back... they always come back :)
Just so I can finish the problem.... Ill step through it like this: c = 45 -9a -3b substitute the "value" of c into the other 2 equations like this: --------------------- 49a+7b+(45 -9a -3b) = 257 49a +7b +45 -9a -3b = 257 40a +4b +45 = 257 40a +4b = 257 -45 40a + 4b = 212 -------------------- 144a+12b+(45 -9a -3b) = 792 144a +12b +45 -9a -3b = 792 135a +9b +45 = 792 135a +9b = 792-45 135a +9b = 747 --------------------- we now have 2 equations to compare: 40a + 4b = 212 135a +9b = 747 lets solve the first one for "b": 40a + 4b = 212 ; divide all by 4 10a+b = 53 b = 53-10a ------------------ Now lets use this "value" of b to solve the other equation by substituting it in: 135a +9(53-10a) = 747 135a + 477 - 90a = 747 135a -90a = 747 -477 45a = 270 a = 270/45 ; reduce this to a=6 a=6 ------------------ we know have a "solid" value for "a". lets use that in our equation for "b". b = 53-10a b = 53-10(6) b = 53-60 b = -7 ------------------- with a "solid" value for "a" AND "b" we can use these in our equation for "c". ------------------------------- c = 45 -9a -3b c = 45 -9(6) -3(-7) c = 45 -54 +21 c = 12 ---------------- a=6, b=-7; c=12 .... if I did it all correctly. So our quadratic equation becomes: f(x) = ax^2 +bx +c f(x) = 6x^2 -7x +12
check to see if that works with our "known" values.... 6(3)^2 -7(3) +12 = 45 6(9) -21 +12 = 45 54 - 9 = 45 45 = 45 ..... that checks --------------------- 6(7)^2 -7(7) +12 = 257 6(49) - 49 +12 = 257 294 - 37 = 257 257 = 257 ....... checks out so far... ------------------- 6(12)^2 -7(12) +12 = 792 6(144) - 84 + 12 = 792 864 - 72 = 792 792 = 792 .... the equation is GOOD!!! lol